Archive for the ‘GRE’ Category

In previous posts, we’ve discussed how to solve systems of equations. But often, you will not be given a simply list of equations. Rather, you will get a chunk of text asking you to find some particular value. For example,

Example 1
Tom and Liz are baking cookies. Tom can bake 10 cookies in an hour and Liz can bake 5 cookies in an hour. Working together, how long will it take them to bake 40 cookies?

Example 2
Dembe is reading a long book. He finishes a quarter of it, then puts it down to take a walk. After he comes back, he reads another 50 pages. Now, he has finished 30% of the book. How long is the book?

Example 3
Raymond is trying to calculate how much he needs to invest today in order to have $2,000,000 in an account in 18 years. He can guarantee an annual return of 15% on any funds invested. How much does he need to invest?

The key to such problems lies in translating the text to mathematical equations. As we will see, solving the actual equations is generally not the hard part; the trick lies in the translation.

The problems above are each examples of different kinds of problems. The first problem is an example of a rate problem, the second one is an example of a ratio problem and the last one is a compound interest problem.

Solving Example 1

Example

Let’s look at our first problem. To determine how long it takes the pair to bake 40 cookies, we would try first figuring out how quickly they bake cookies. Let’s try to figure out how many cookies they would bake in an hour. Well, during that hour, Tom would bake 10 cookies and Liz would bake 5 cookies. So together, they would bake 15 cookies.

Now, we ask: if someone could bake 15 cookies in an hour, how long would it take them to bake 40 cookies? To answer this, we use an old equation:

    \[d = rt\]

where d is the total number of cookies baked, r is the number of cookies baked each hour, and t is the amount of hours spent baking cookies. Turning to our problem, we know that their combined rate is 15 cookies per hour, and we want to know how long it takes to bake 40 cookies. Therefore, we set up the following equation:

    \[40 = 15t\]

and solve for t. We learn, by dividing both sides by 15, t = \frac{8}{3}. So it takes them \frac{8}{3} hours to do so (or, in other words, 2 hours and 40 minutes).

This kind of problem often involves some kind of task (eg baking cookies) or distance (eg how far one runs). The key is to use what you know about the outcome (d), rate (r), or time (t) to solve for the unknown quantity. In this case, the problem gave us the rate and outcome, and we solved for the time it took to get that outcome.

Solving Example 2

Example

The key here is setting up the right equation. Let’s use our variables. Let p = the number of pages in Dembe’s book. We know that, after reading .25p + 50 pages, Dembe has finished .3p pages (i.e. 30% of the book). Thus, we get

    \[.25p + 50 = .3p\]

Now that we have this equation, it is easy to solve for the value of p. We subtract to get:

    \[50 = .05p\]

    \[p = 1000\]

Thus, we conclude that Dembe’s book was 1,000 pages long.

Solving Example 3

Example

Again, let’s use our variables. Let x = the amount of money Raymond invests today. Then, in 18 years, with an annual rate of 15%, Raymond will have x(1.15)^18 in that account. We want to make that amount equal to 1,000,000. Thus, we get the equation:

    \[x(1.15)^{18} = 1,000,000\]

And now, we can solve for x by dividing by 1.15^{18}:

    \[x \approx 80805.\]

Thus, Raymond needs to invest about $80,805 today.

Now there is no cookie-cutter recipe that will handle all word problems. But in general, thinking about what the words mean and translating them into mathematical formulas carefully will work.

Practice Problems:

  1. Felipe left for his morning jog with a water bottle that was three-quarters full. On his route, he passed by a high-tech water fountain and added precisely 300 milliliters of water to his bottle (the bottle still was not full). Then, he continued on his run. At the end of it, he drank two-thirds of the water then in his bottle before heading home. Let w be the maximum capacity of Felipe’s water bottle (in milliliters). Write an expression for how much water was left in his water bottle when he reached his home. If he had exactly 200 milliliters of water when he reached home, what was the capacity of his water bottle?
    Answer
  2. The kindergarteners are trying to decide which shape is best. Every kindergartener likes only one shape. 3/5 of the class is tall and the rest are short. Of the tall students, 1/6 like circles, 2/3 like squares, and everyone else likes trapezoids. Suppose 1/2 of the class likes circles. What proportion of the short students like circles?
    Answer

  3. Evan is a distance runner. He runs at a constant rate of 10 miles per hour for as many hours as you like. How long will it take him to finish a marathon (26.2 miles)?
    Answer

 


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We’ve talked before about solving systems of linear inequalities. There, we simply graph the relevant equations to find where the shaded regions all intersect. We do the same thing for systems of quadratic inequalities. Let’s look at an example:

    \[\begin{cases} y \geq  & x^2  \\  y \leq  & x^2 + 10 \end{cases}\]

We begin by graphing the two equations:

Then, we shade in the regions where each inequality applies. So, for example, where will y \geq x^2? Clearly this will be true in the region above the curve. (And if you are unsure, you can always just try testing a point on each side of the curve). We do the same for the second equation to get:

And as before, the solutions will be any point in the area where all of the shaded regions intersect.

Now, we can also have systems of inequalities that combine linear and quadratic inequalities:

    \[\begin{cases} y \geq & x^2 \\ y \leq & 2x + 5 \end{cases}\]

We follow the same method, graphing the functions:

Shading the relevant regions:

And identifying where shaded regions intersect:

And that’s how we handle systems of inequalities for quadratic equations.

Practice Problems

For each of the following systems of quadratic inequalities, graph their solutions. (Some may not have any solutions at all).

  1.  

        \[\begin{cases} y <& x^2 - 5 \\ y >& x - 5 \end{cases}\]

    Answer
  2.  

        \[\begin{cases}y <& x^2 + x \\ x >& 5 \\ y <& 5\\ y >& x \end{cases}\]

    Answer
  3.  

        \[\begin{cases}y >& 3x^2\\ y <& 0 \end{cases}\]

    Answer

 


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Now, sometimes questions will give you a set of inequalities, rather than a set of equations. For example,

    \[\begin{cases} y \leq & 3x + 4 \\ y > & 2x - 3 \end{cases}\]

In these problems, we are not looking for specific solutions for x and y. This is because there are many solutions to the above equations. For example, (0,0) is a solution and so is (1, 6). Rather, we want to characterize, in a general way, what the complete set of solutions looks like.

The easiest way to do this is by graphing the above equations. If we graph them, we get:

Now, we want to determine where the inequalities hold. So suppose we’re interested in the first inequality, y \leq 3x + 4. Where will this be true?

It will be true below (and on) the line y = 3x + 4. We can mark this on our diagram by shading in that region:

And typically, we use a solid line to designate \leq, \geq.

Now, what about the second inequality, y > 2x - 3? Typically, we use a dotted line to designate <, >. So, we get:

And now we can go to our graph and we ask: where is y greater than 2x - 3? We have this line showing where y = 2x - 3 so y must be greater than 2x - 3 whenever we are above this line. (Alternatively, if you’re unsure which side of the line satisfies the inequality, pick a point on each side and test whether it satisfies the inequality).

Now, where will both of these inequalities be true? Well recall that the red shading indicates the first inequality is true and the blue shading indicates the second inequality is true. So the inequalities are both true at any point that is shaded both red and blue. So the solutions to these inequalities are just the points in this region that is shaded both blue and red. In other words:

Now sometimes your equations may not seem easy to graph. For example, you might get

    \[2x - \frac{y}{9} = 20\]

But with some algebra, we can always convert it into slope-intercept form:

    \[y = 18x - 180\]

and then we can graph it more easily. 

Finally, sometimes your inequalities will have no solution. For example:

    \[\begin{cases} y <& x \\ y > & x + 1 \end{cases}\]

has no solution, as can be verified by looking at the graph:

and noticing that there is no point that falls into both shaded regions. In this case, we just say that the system has no solution.

Finally, it is important to know how to handle equations like x \geq 3. To address these cases, we need to know how to graph a function like x = 3. To graph that function, we just draw a vertical line at any point with an x-coordinate of 3, as seen here:

And to tell which side of the line to shade in, we can just try a point on each side. (4, 0) satisfies x \geq 3 whereas (2, 0) does not. Thus, we shade in the side to the right, and we use a solid line because this is a weak inequality (i.e. includes the possibility x = 3).

Practice Problems

Graph the solutions to the following:

  1.  

        \[\begin{cases} 2x - y <& 5 \\ 2y + x >& 5 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 3y - 2x \geq & 2 \\ 2x - y < & 100 \end{cases}\]

    Answer
  3.  

        \[\begin{cases} 2y + 2x \geq & 4 \\ x - y \geq & 1 \end{cases}\]

    Answer

 


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In a previous post, we have seen how to use the quadratic formula to solve quadratic equations. Here, we will discuss another method: factoring.

The basic idea behind factoring is the same idea that drives many military campaigns: divide and conquer. A quadratic equation (like x^2 + 5x + 6 = 0) is complicated. Finding its solutions is hard. But if we break it down into its component parts:

    \[x^2 + 5x + 6 = (x+3)(x+2) = 0\]

then it is easy to find its solutions! After all, the only way for (x+3)(x+2) to be equal to 0 is if (x+3) = 0 or (x+2) = 0 or both are equal to 0. Otherwise, we would have two non-zero numbers multiplied by each other, which always yields a non-zero number. 

And the only values of x that make x+3 or x+2 equal to 0 are -3 and -2. Thus, x = -3, -2 are the only solutions of (x+3)(x+2) = 0. And since (x+3)(x+2) = x^2 + 5x + 6, we conclude that x = -3, -2 are the only solutions of x^2 + 5x + 6 = 0.

Now, let’s set out the general steps for factoring a quadratic equation:

Method of Factoring

  1. Ensure 0 is on one side of the equation
  2. Factor our quadratic expression as a product of two linear expressions
  3. Find the x-values that make each linear expression equal to 0

Example
Suppose we start with: x^2 + 2x - 6 = 10x + 3

  1. Ensure 0 is on one side of the equation
    Remember that we can add/subtract/multiply/divide as we like, provided that we do it to both sides. (Note, though, that we are not allowed to divide by x or 0. Dividing by x eliminates a solution and dividing by 0 is just prohibited by the rules of arithmetic).Let’s subtract 10x + 3 from both sides. Then we get:

        \[x^2 - 8x - 9 = 0.\]

  2. Factor our quadratic expression as a product of two linear expressions
    Linear expressions are just expressions that have a degree of 1 or lower. For example, things like x - 2 and 7x + 49 and -2 are all linear expressions.
    In the next section, we will further discuss how to find the right linear expressions. Here, we show how we might do that for the quadratic expression x^2 - 8x -9.
    We know that we are ultimately looking for something like:

        \[(ax + b)(cx + d) = x^2 - 8x - 9\]

    where a, b, c, d are all constants. This is because ax + b is the general form for any linear expression. Let’s suppose we have already found the right a, b, c, d. Then, multiplying things out, we get:

        \[(ax+b)(cx+d) = (ax)(cx) + bcx + axd + bd = x^2 - 8x - 9\]

        \[\implies ac(x^2) + (bc + ad)x + bd = x^2 - 8x - 9\]

    Now, on the left hand side of the equation, we have that the only coefficient of x^2 is ac. On the right hand side, we know that the coefficient of x^2 is 1. Therefore, ac = 1 (since otherwise, the two sides of the equation would not be equal, since they would have different amounts of x^2).Now, that means we could have a = 1, c = 1 or a = -1, c = -1 or a = \frac{1}{2}, c = 2 and many more possibilities. Just to keep things simple, though, let’s try assuming that a = 1, c = 1 and see if that leads us to a solution. (If it does not, we can always revisit this step and try something else).So suppose a = 1, c = a. Then, we get:

        \[(ac)x^2 + (bc+ad)x + bd = x^2 + (b + d)x - bd = x^2 - 8x - 9.\]

    Now, we want to find some values of b, d such that:

        \[\begin{cases} b+d =& -8 \\ bd =& -9 \end{cases}\]

    Now, we will generally solve these problems by listing all the possible integers such that bd = -9. Then, we will look at them and see if any of those combinations will add up to -8. If we cannot find an integer solution, then your best bet is to give up on this method and just use the quadratic formula.
    Here are all the possibilities for bd = -9.You can construct such a list by going through the factors of 9. For every factor of 9 (e.g. 1) there is some integer (in this case, 9) so that when they are multiplied together, they equal 9 (1 \times 9 = 9). And to deal with the negative sign, simply alternate which of the two factors (whether 1 or 9) is negative.Running through the table, we get:

    Thus there are two possibilities: b = 1, d = -9 or b = -9, d = 1. It doesn’t matter which one you choose. Let’s pick the first one.
    Then, by plugging in our values for b, d, we get

        \[(x + 1)(x - 9) = x^2 - 9x + x - 9 = x^2 - 8x - 9\]

    which is exactly what we wanted.

  3. Find the x-values that make each linear expression equal to 0
    Then, x + 1 = 0 implies that x = -1. And x-9 = 0 implies x = 9. Thus, x = -1, 9. Those are the solutions to x^2 - 8x - 9 = 0.

Factoring into Linear Expressions

The trickiest step when using factoring to solve a quadratic equation is finding the linear expressions to multiply together. Here, we will only deal with quadratic equations of the form:

    \[x^2 + bx + c = 0\]

where b, c are constants. In other words, we assume that the coefficient of x^2 is 1.

Now you can factor equations that have other coefficients for x^2. For example,

    \[2x^2 - 5x - 3 = 0\]

can be factored as

    \[(2x + 1)(x - 3) = 0\]

, thereby yielding the solutions x = \frac{-1}{2}, 3. But dealing with these cases is harder, so you are better off, generally, just applying the quadratic formula.

Here are the steps for factoring our quadratic equation into linear expressions:

  1. Write out the factors of our constant, c
  2. Write out the pairs of factors that multiply together to get c
  3. Go through them to see which pairs of factors add up to b
  4. (If time) Check our solution

Let’s look at another example. Suppose we have: x^2 - 5x + 6 = 0. We want to factor this into an expression that looks like: (x + m)(x + n) where m, n are integers. Now, let’s follow our steps.

Factoring into Linear Expressions

  1. Write out the factors of our constant, c
    6 has the factors 1, 2, 3, and 6.
  2. Write out the pairs of factors that multiply together to get c
    The easiest way to do this is just to go from least factor to the greatest factor in the first column, and then write down the number needed in the second column so that their product is equal to 6. Here, we have:

    And don’t forget that combinations with negative numbers also count:

    Now, this may look like a lot of combinations to try. But note that the first row of each table looks like the fourth row, just with the order of the columns flipped. Since it doesn’t matter which order the numbers are in (since m + n = n + m), we don’t need to check both rows. Similarly for the second and third row. Thus, we only need to check:
  3. Go through them to see which pairs of factors add up to b
    Now, we just check whether any of these combinations add up to -5. And -2 + -3 = -5. Thus, we have our answer: (x-2)(x-3) = x^2 -5x + 6.
  4. (If time) Check our solution
    If you have the time, it is highly recommended that you check your solution since it only takes a second to do. And if your solution is wrong, that’s valuable information. In our case:

        \[(x-2)(x - 3) = x^2 - 2x - 3x + 6 = x^2 - 5x + 6\]

    so our solution is correct.

Practice Problems

Solve the following quadratic equations via factoring:

  1.  

        \[x^2 - 6x = 0\]

    Answer
  2.  

        \[y^2 +3y -18 = 0\]

    Answer
  3.  

        \[x^2 - 6x + 9 = 0\]

    Answer
  4.  

        \[x^2 + 4x + 7 = -2x -1\]

    Answer
  5.  

        \[2x^2 - 14x + 24 = 0\]

    Answer

 


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In this post, we will discuss the most general way to solve a quadratic equation: the quadratic formula. If you only have enough time to learn one way of solving a quadratic equation, you should just memorize this formula (and skip this post on factoring) since this strategy is guaranteed to work (whereas sometimes factoring works well and sometimes not). Here is how the method works:

Quadratic Formula Method
Suppose we have a quadratic equation of the form: ax^2 + bx + c = 0. Then, the solutions are:

    \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Note that (provided b^2 - 4ac \neq 0), this means that we will have 2 solutions for our equation: one where we add the square root and another where we subtract it. And in general, quadratic equations do have 2 solutions.

This is a very powerful formula and basically allows one to solve any quadratic equation, provided one is willing to learn the formula and, sometimes, do a little bit of manipulation to the initial equation.

Here is an example problem:

Example
Find the solution(s) to:

    \[\frac{9x^2}{7} + 3x - \frac{34}{7} = 2x - 2\]


This is a somewhat ugly equation so before proceeding, we do some multiplication to get rid of the fractions. Multiplying by 7, we get:

    \[9x^2 + 3x - 34 = 14x - 14\]

And before using our formula, it is crucial that all the terms are on one side, and that the other side is equal to 0. Thus, we subtract 14x - 14 to get:

    \[9x^2 + 3x - 34 - (14x - 14) = 14x - 14 - (14x-14)=0\]

    \[\implies 9x^2 + 3x - 34 - 14x + 14 = 0\]

    \[\implies 9x^2 - 11x - 20 = 0\]

Now, we can apply the formula to get:

    \[x = \frac{-(-11) \pm \sqrt{121 - 4(9)(-20)}}{2(9)} = \frac{11 \pm \sqrt{841}}{18}\]

    \[= \frac{11 \pm 29}{18}\]

Thus, x = \frac{20}{9}, -1.

Remembering the Formula
Using the quadratic formula is generally straightforward. The tricky part is remembering the formula itself. One way to do so is via a song. If you are familiar with the tune of “Brother Jacques,” you could sing the following:

Practice Problems
Find the solutions to the following quadratic equations:

  1.  -6x^2 - 24x + 3 = 30

    Answer

  2.  x^2 - \frac{2}{9}x = x

    Answer

  3.  x^2 = x
    Answer

  4.  x = 4x^2 - 30
    Answer

  5.  3x^2 = 9
    Answer

 


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In previous posts, we have discussed how to solve systems of linear equations. Now, we will talk about a bigger family of equations: the polynomials. Formally, the polynomials are all the equations of the form:

    \[a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0\]

where a_i is some constant (for i = 0, 1, ..., n) and x is a variable. Basically, polynomials are combinations of constants and the positive powers of x multiplied by any coefficient.

We say that the highest exponent of a variable in a polynomial is its degree. So for example,

    \[x^{90} + 2 = 0 \textrm{ has degree 90}\]

    \[x^2 + 2x - 8 = 0 \textrm{ has degree 2}\]

    \[0x^{90} + x^2 + 2x - 8 = 0 \textrm{ has degree 2}\]

since if we simplify the last equation, the x^{90} term goes away and we just get x^2 + 2x - 8 = 0 which we earlier noted has degree 2.

We say equations with just constants have degree 0. So for example,

    \[10 - 8 = 2 \textrm{ has degree 0}\]

Now, some polynomials play a distinctive role in mathematics and so they get special names. Polynomials with degree 2 (i.e. 2 is the highest power of any variable) are called quadratic equations. Here are some examples of quadratic equations:

    \[x^2 + x = 0\]

    \[3x^2 - 2x + 5 = 0\]

    \[\frac{x^2}{9} - x + 3 = x^2 - 2x + 8\]

And while the last equation may not look like our definition of a polynomial, we can see how by subtracting x^2 -2x + 8 from both sides, we can put it into that form:

    \[\frac{-17x^2}{9} + 3x - 5 = 0\]

Polynomials with degree 3 are called cubic. Here are some examples of those:

    \[x^3  = 0\]

    \[9x^3 + x^2 - 2x + 1 = 0\]

    \[\frac{x^3}{9} - x + 3 = x^2 - 2x + 8\]

In the next few posts, we will discuss two strategies for solving quadratic equations: via the quadratic formula or via factoring.

Practice Problems

Find the degree of the following equations:

  1. \frac{9x^3}{x} - 9x  = 0, where x \neq 0

    Answer

  2. 9x^4 - 20x^9 = 0

    Answer

  3. x - 2x + 3x^2 - 4x + x^5 = 0

    Answer


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In the previous two sections, we’ve solved sets of two or even three linear equations. Here, we will do some examples of systems with four linear equations. The general strategy will be the same as before: we apply our method (either elimination or substitution) to our sets of equations in order to get rid of one equation. We repeat this until we’re left with just two or three equations, whereupon we know how to solve the problem. Also note that it takes n distinct equations to solve a system of n linear equations. (Two equations are distinct if you cannot obtain either one from the other).

Example (Elimination Method for 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to eliminate.
    Let’s pick a.
  2. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
    Now, if we look at our equations, we see that equation 4 does not have any a’s in it. So no matter what constant we multiply the equation by, we will not get a term with a multiplied by some given coefficient. If we have an equation like this, we simply set it to the side and use it again later. This is because, recall, the point of this method is to eliminate all occurrences of a given variable, thereby simplifying our problem. Equation 4 has already eliminated all occurrences of a, so no further manipulation of the equation is needed.Now, we get:

        \[\begin{cases} 12a + 6b + 6c + 3d - 3e &= 24 \\ 12a - 12b + 36c + 12d &= 48 \\ 12a + 12b - 4c + 4d - 4e &= 24 \\ 12a - 12b + 12c - 12d + 12e &= -12 \end{cases}\]

    And we still have our equation 4:

        \[3b - c + 2d - 3e =& -12\]

  3. Subtract the equations from one another, thereby eliminating our variable.
    Note that this only applies to the equations that have a in them. And remember to use all of the equations that include a (but we only need to use each equation once).Then, we get:

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ -24b + 40c + 8d + 4e &= 24\\ -24b -16c + 16d - 16e &= 36 \\ 3b - c + 2d - 3e &= -12\end{cases}\]

  4. Now, we should have four equations with at most four different variables. Pick another variable to eliminate.
    Let’s eliminate b.
  5. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ 18b - 30c - 6d - 3e &= -18\\ 18b +12c - 12d + 12e &= -27 \\ 18b - 6c + 12d - 18e &= -72\end{cases}\]

  6. Subtract the equations from one another, thereby eliminating our variable.

        \[\begin{cases} -3d &= -6\\ -42c + 6d -15e &= 9\\ 18c - 24d + 30e &= 45 \end{cases}\]

  7. Now, we should have three equations with at most three different variables.
    And we do.
  8. We can now solve this problem with the steps detailed in our previous post on the elimination method.

Example (Substitution Method with 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to substitute for.
    Let’s choose a.
  2. Use one of the lines of the equation to express that variable in terms of the other variables.
    Let’s use the second equation. Then, we get:

        \[a = 4 + b - 3c - d\]

  3. Substitute into the other equations.
    Since we used the second equation to get our equation for a, we now substitute for a in the first, third, fourth, and fifth equations. We then get (after simplifying):

        \[\begin{cases} 6b -10c -3d - e &= -8\\ 6b -10c -2d - e &= -6\\ 3b - c + 2d - 3e &= -12\\ -2c - 2d + e &= -5 \end{cases}\]

  4. Now, we should have four equations with at most four different variables.
    And this is true.
  5. Pick another variable to substitute for.
    Let’s pick e.
  6. Use one of the remaining lines of the equation to express that variable in terms of the other variables.
    Let’s use the first line. Then, we get

        \[e = 6b - 10c - 3d + 8\]

  7. Substitute into the other equations.
    Since we used the first equation, we need to plug in our formula for e into the second, third, and fourth equations. We then get (after simplifying):

        \[\begin{cases} d &= 2\\ -15b + 29c + 11d &= 12\\ 6b - 12c - 5d &= -13 \end{cases}\]

  8. Now, we should have three equations with at most three different variables.
    And we do.
  9. We can now solve this problem with the steps detailed in our previous post on the substitution method.

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In a previous post, we talked about solving systems of linear equations with the method of elimination. Here, we talk about the method of substitution.

Essentially, for this method we want to get an equation that expresses one variable in terms of the others. For example, suppose we start with the following system:

Example

    \[\begin{cases} x + 2y &= 3 \\ 2x - 3y &= 4 \end{cases}\]

We can re-write the first equation as:

    \[x = -2y + 3\]

Thereby giving us x in terms of y and some constants. We can now plug this into the second equation, giving us:

    \[2(-2y + 3) - 3y = 4\]

And simplifying, we get:

    \[-4y + 6 - 3y = -7y + 6 = 4\]

    \[\implies y = \frac{2}{7}\]

Now that we have the value of y, we can plug it into one of our equations to get the value of x:

    \[x + 2(\frac{2}{7}) = 3\]

    \[\implies x = \frac{17}{7}\]

In general, the steps of our method are as follows. Suppose we have some system of equations:

    \[\begin{cases} ax + by &= A \\ cx + dy &= B\end{cases}\]

where A, B are constants.

Substitution Method

  1. Pick one variable to substitute.

For our example, let’s pick x to substitute.

  1. Define that variable in terms of the other variables.

We can use either equation to define the value of x. Let’s use the first. Then, we get:

    \[x = \frac{A-by}{a}\]

  1. Substitute that variable into the remaining equation to solve for one of the variables.

Since we used the first equation to define x, we need to plug it into the second equation. Then, we get:

    \[cx + dy = \frac{A-by}{a}c + dy = B\]

    \[\implies y = \frac{B - \frac{A-by}{a}c}{d}\]

  1. Now that you have one of the variables, use the equations to find the other.

Just plug in your value for y into ax + by = A to find the value of x.

Now, as before, we can also use this method for systems of 3 equations. There, we can turn that problem into a system of 2 equations as follows.

Example

Suppose we have the following system of linear equations:

    \[\begin{cases} 3x + 4y - z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z &= 2 \end{cases}\]

Now, we pick a variable that we want to substitute. Suppose we want to substitute x first. The key is that we can pick any of the three equations to define x in terms of the other variables, y and z. Then, we use that equation to substitute for x in the other two equations.

Suppose we want to use equation 2 to define x. Then, we get:

    \[x = 4 - 2y - 2z\]

Then, since we used equation 2 to define x, we need to substitute for x in equations 1 and 3 (i.e. the other equations). We then get:

    \[\begin{cases} 3(4-2y-2z) + 4y - z &= 8 \\ 2(4-2y-2z) - 3z &= 2 \end{cases}\]

Simplifying, we get:

    \[\begin{cases}-2y - 7z &= -4 \\ -4y - 7z &= -6 \end{cases}\]

Which we can solve via our standard substitution methods.

Practice Problems

Solve the following systems of equations using the method of substitution.

  1.  

        \[\begin{cases} 7x - 2y &= x + 3 \\ x + 2y &= 0 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 7x - 2y &= 26\\ 2x + 2y &= 15 \end{cases}\]

    Answer

  3.  

        \[\begin{cases} x + 3y &= 15\\ x &= -y + 5 \end{cases}\]

     

    Answer
  4.  

        \[\begin{cases} 4x + 2y &= 3x - 3y + 3\\ 2x - y + 2 &= x - 2y \end{cases}\]

     

    Answer
  5.  

        \[\begin{cases} x + 4y - 3z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z = 7 \end{cases}\]

     

    Answer

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In this post, we will talk about how to solve a system of linear equations via the method of elimination. Now, you will not need to know both this method and the method of substitution that we discuss in the next post. You do need to be able to solve a system of linear equations, but either method suffices for that purpose. However, for some problems, one or the other method may be easier. So if you have the time, it is worth learning both.

Now suppose we have a system of equations as follows:

Example

    \[\begin{cases} x + 2y &= 20 \\ x + y &= 8 \end{cases}\]

We could try to solve this system by guessing values of x and y, and plugging them in to see if they work. But that would take a long time. Instead, note that if we subtract the second equation from the first, we get:

    \[x + 2y - (x+y) = 20 - 8\]

Or, in other words,

    \[y = 20-8 = 12\]

Now that we have found the value of y, we can plug it back in to find the value of x:

    \[x + (12) = 8\]

    \[\implies x = -4\]

So our solution is (-4, 12).

In general, the strategy for solving equations by elimination is to try and get some variables to cancel each other out. We ultimately want one line of the equation to have just one variable in it, for then it is easy to calculate the value of that variable. Once we have that value, we simply plug it back into an earlier equation to get the value of the remaining variable(s).

Let’s look at another example:

Example

    \[\begin{cases} 3x + y &= 8 \\ x + 2y &= 9 \end{cases}\]

Here, it is not useful to simply subtract the second equation from the first. Doing so yields:

    \[2x - y = -1\]

which is no more helpful than what we started with.

Instead, we need to make a decision: do we want to eliminate x or y? It does not matter too much which one you choose (at most you can save yourself a few lines of arithmetic). Let’s say we want to eliminate y. Then, we need to manipulate our equations so that the y variable will be eliminated. We see that in our second equation, y has a coefficient of 2 but in the first equation, y only has a coefficient of 1.

Subtracting these equations, then, will get us an equation that includes y. To get the y’s to cancel out, we need y to have the same coefficient in both equations. So let’s multiply the first equation by 2. (We could have also divided the second equation by 2, but fractions are more annoying to work with). Then, we get

    \[\begin{cases} 6x + 2y &= 16 \\ x + 2y &= 9 \end{cases}\]

And now, we can subtract the second equation from the first to get:

    \[5x = 7\]

    \[\implies x = \frac{5}{7}\]

Plugging this back into one of our initial equations, we get:

    \[x + 2y = \frac{5}{7} + 2y = 9\]

    \[\implies 2y = \frac{58}{7}\]

    \[\implies y = \frac{29}{7}\]

So our solution is (\frac{5}{7}, \frac{29}{7}).

In general, elimination problems will look like the following:

    \[\begin{cases} ax + by &= A \\ cx + dy &= B\end{cases}\]

where A, B are constants. Let’s lay out the steps for using the method of elimination in general:

Elimination Method

  1. Pick either x or y to eliminate.

For our example, suppose we want to eliminate x.

  1. Force x to have the same coefficient in both equations.

To do that here, we multiply the first equation by \frac{c}{a} to get:

    \[cx + \frac{bc}{a}y = \frac{cA}{a}\]

(The fact that we chose to manipulate the first equation is unimportant. We could have instead multiplied the second equation by \frac{a}{c} to get similar results.)

  1. Finally, we subtract the second equation from the first (or the first from the second, you can pick whichever is easier to calculate).

Here, we get:

    \[(d-\frac{bc}{a})y = \frac{cA}{a} - B\]

    \[\implies y = \frac{\frac{cA}{a} - B}{d-\frac{bc}{a}}\]

And while this looks rather complicated, recall that a, b, c, d, A, B are all just constants. So finding the value of y is now just a matter of plugging in some numbers into a calculator.

  1. Once we know the value of y, we can just plug it in to get the value of x.

Here, x = \frac{B-dy}{c}.

Now, we can also use the method of elimination to solve systems of 3 equations. Here, the trick is to turn our system of 3 equations into a system of 2 equations. This is done by picking some variable to eliminate, and then applying (roughly) the above method to eliminate it. Then, we apply the above steps method in order to solve the remaining system of 2 equations. Here is an example of such a problem.

Example

    \[\begin{cases} 3x + y - z &= 7 \\ 2x + 4y + 3z &= 24 \\ x - 2y + 2z &= 9\end{cases}\]

Suppose we want to eliminate the variable z. Then, we multiply our equations by constants so that z will have the same coefficient in every equation. Since we want to avoid working with fractions, let’s try to make z have the coefficient of 6 in every equation. Then, we multiply the first equation by -6, the second one by 2, and the third one by 3 to get:

    \[\begin{cases} -18x -6 y +6 z &= -42 \\ 4x + 8y + 6z &= 48 \\ 3x - 6y + 6z &= 27\end{cases}\]

Now, we need to cancel out the z’s. How do we do this? Now, as before, there are many ways to cancel out the z’s. The important thing is that we use all three equations in this process. (For an explanation of why this is important, see this post. Intuitively, if you don’t use all three equations, you are losing information). So we cannot simply subtract the second equation from the first and then subtract the first equation from the second.

In our case, we will simply subtract the second equation from the first, and then subtract the third equation from the second. This yields:

    \[\begin{cases} -22x -14y &= -100 \\ x + 14y &= 21 \end{cases}\]

And we can solve this using the normal method of elimination discussed earlier.

Like any mathematical ability, really becoming proficient with the method of elimination requires practice. Here are some problems to practice on.

Practice Problems

Solve the following systems of equations using the method of elimination

  1.  

        \[\begin{cases} 8x - y &= 2x + 6 \\ 2x + y &= 0 \end{cases}\]

    Answer
  1.  

        \[\begin{cases} 9x - 2y &= 26\\ 3x + 2y &= 13 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 2x + 3y &= 15 \\ x &= -y + 5 \end{cases}\]

    Answer
  3.  

        \[\begin{cases} 2x + 2y &= 3x - 3y + 8\\ 3x - y + 2 &= x - 2y \end{cases}\]

     

    Answer

 


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