Archive for the ‘GRE’ Category
Here are all the posts related to solving equations:
 Solving Equations: Some Prerequisites
 Solving Equations: An Introduction
 Solving Equations: Linear Equations
 Solving Equations: Polynomials
 Equations: Word Problems
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,
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?
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?
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.
Problem 1
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:
where is the total number of cookies baked, is the number of cookies baked each hour, and 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:
and solve for . We learn, by dividing both sides by 15, So it takes them 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 (), rate (), or time () 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.
Problem 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?
The key here is setting up the right equation. Let’s use our variables. Let = the number of pages in Dembe’s book. We know that, after reading pages, Dembe has finished pages (i.e. 30% of the book). Thus, we get
Now that we have this equation, it is easy to solve for the value of :
Thus, we conclude that Dembe’s book was 1,000 pages long.
Problem 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?
Again, let’s use our variables. Let = the amount of money Raymond invests today. Then, in 18 years, with an annual rate of 15%, Raymond will have in that account. We want to make that amount equal to 1,000,000. Thus, we get the equation:
And now, we can solve for by dividing by :
Thus, Raymond needs to invest about $80,805 today.
Now there is no cookiecutter recipe that will handle all word problems. But in general, thinking about what the words mean and translating them into mathematical formula carefully will work.
Practice Problems:
 Felipe left for his morning jog with a water bottle that was threequarters full. On his route, he passed by a hightech 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 twothirds of the water then in his bottle before heading home. Let be the maximum capacity of Felipe’s water bottle. 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?
 The kindergarteners are trying to decide which shape is best. Every kindergartener likes only one shape. 3/5 of the class is female and 2/5 of the class is male. Of the females, 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 males like circles?
 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)?
We’ve talked before about solving systems of linear inequalities (%TODO insert hyperlink). 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:
We begin by graphing the two equations:
[graph of the two]
Then, we shade in the regions where each inequality applies. So, for example, where will ? 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:
[graph of the two with shaded regions]
And as before, the solutions will be any point in the area where all of the shaded regions intersect.
[graph of the two with shaded regions and an arrow to the solution space]
Now, we can also have systems of inequalities that combine linear and quadratic inequalities:
We follow the same method, graphing the functions:
[graph]
Shading the relevant regions:
[graph with shaded in regions]
And identifying where shaded regions intersect:
[graph with shaded in regions and arrow pointing to the solution]
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).
 y < x^2  5; y > x  5
 y < x^2 + x; x > 5; y < 5; y > x
 y > 3x^2; y < 0
Now, sometimes questions will give you a set of inequalities, rather than a set of equations. For example,
In these problems, we are not looking for specific solutions for and . This is because there are many solutions to the above equations. For example, is a solution and so is 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:
[Graphic with just the lines, one blue and one red]
Now, we want to determine where the inequalities hold. So suppose we’re interested in the first inequality, Where will this be true?
It will be true above (and on) the line We can mark this on our diagram by shading in that region
[Graphic with shaded blue region]
Typically, we use a solid line to designate .
Now, what about the second inequality, ? Again, we can go to our graph and we ask: where is greater than ? We have this line showing where so must be greater than 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).
Typically, we use a dotted line to designate . So, we get:
[Graphic with shaded blue region, shaded red region, and red dotted line]
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.
Now sometimes your equations may not seem easy to graph. For example, you might get
But with some algebra, we can always convert it into slopeintercept form:
and then we can graph it more easily. (For more information on graphing functions, see here: %TODO Add page)
Finally, sometimes your inequalities will have no solution. For example:
has no solution, as can be verified by looking at the graph:
[Graphic with shaded regions]
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 To address these cases, we need to know how to graph a function like To graph that function, we just draw a vertical line at any point with an xcoordinate of 3, as seen here:
[graph of x = 3]
And to tell which side of the line to shade in, we can just try a point on each side. satisfies whereas does not. Thus, we shade in the side to the right, and we use a dotted line because this is a strict inequality (i.e. rules out the possibility ).
[Graph of x = 3 with (4, 0) and (2, 0) and shade and dotted line]
And to tell which side of the line to shade in, we can just try a point on each side. satisfies whereas does not. Thus, we shade in the side to the right, and we use a dotted line because this is a strict inequality (i.e. rules out the possibility ).
[Graph of x = 3 with (4, 0) and (2, 0) and shade and dotted line]
Practice Problems
Graph the solutions to the following:
 2x  y < 5; 2y + x > 5
 3y  2x > 2; 2x  y < 100
 2y + 2x > 4; x  y > 1
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 idea behind factoring is that we take a quadratic equation, say and we break it down into two expressions that are multiplied together. In our case,
Then, we have that And the only way for to be equal to 0 is if or or both are equal to 0. Otherwise, we would have two nonzero numbers multiplied by each other, which always yields a nonzero number.
Thus, are the only solutions of And since , we know that are the only solutions of
Now, let’s lay out the steps of factoring a quadratic equation:
 Ensure 0 is on one side of the equation
 Factor our quadratic expression as a product of two linear expressions
 Find the values that make each linear expression equal to 0
Suppose we start with:
 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 or 0. Dividing by eliminates a solution and dividing by 0 is just prohibited by the rules of arithmetic).
Let’s subtract from both sides. Then we get:
 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 and and 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
We know that we are ultimately looking for something like:
where are all constants. This is because is the general form for any linear expression. Let’s suppose we have already found the right Then, multiplying things out, we get:
Now, on the left hand side of the equation, we have that the only coefficient of is On the right hand side, we know that the coefficient of is 1. Therefore, (since otherwise, the two sides of the equation would not be equal, since they would have different amounts of ).
Now, that means we could have or or and many more possibilities. Just to keep things simple, though, let’s try assuming that 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 . Then, we get:
Now, we want to find some values of such that:
Now, we will generally solve these problems by listing all the possible integers such that Then, we will look at them and see if any of those combinations will add up to . 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
b d
9 1
3 3
1 9
1 9
3 3
9 1
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 (). 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:
b d b+d
9 1 8
3 3 0
1 9 8
1 9 8
3 3 
9 1 8
Thus there are two possibilities: or It doesn’t matter which one you choose. Let’s pick the first one.
Then, by plugging in our values for , we get
which is exactly what we wanted.
 Find the xvalues that make each linear expression equal to 0
Then, implies that . And implies Thus, Those are the solutions to
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:
where are constants. In other words, we assume that the coefficient of is 1.
Now you can factor equations that have other coefficients for For example,
can be factored as
, thereby yielding the solutions . 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:
 Write out the factors of our constant,
 Write out the pairs of factors that multiply together to get
 Go through them to see which pairs of factors add up to
 (If time) Check our solution
Let’s look at another example. Suppose we have: We want to factor this into an expression that looks like: where are integers. Now, let’s follow our steps.
 Write out the factors of our constant,
has the factors 1, 2, 3, and 6.
 Write out the pairs of factors that multiply together to get
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:
m n
1 6
2 3
3 2
6 1
And don’t forget that combinations with negative numbers also count:
m n
1 6
2 3
3 2
6 1
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 ), we don’t need to check both rows. Similarly for the second and third row. Thus, we only need to check:
m n
1 6
2 3
1 6
2 3
 Go through them to see which pairs of factors add up to
Now, we just check whether any of these combinations add up to 5. And . Thus, we have our answer: .
 (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:
so our solution is correct.
Practice Problems
Factor the following quadratic equations:
 x^2  6x = 0
 x^2 +3x 18 = 0
 x^2  6x + 9 = 0
 x^2 + 4x + 7 = 2x 1
 2x^2  14x + 24 = 0
Solutions
 x(x6)
 (x+6)(x3)
 (x3)^2
 (x+4)(x+2)
 (x3)(x4)
We said earlier that we would only deal with factor equations where the coefficient of is 1. But we can easily turn this equation into that kind of problem. Simply divide both sides of the equation by 2 to get:
which we can solve via our factoring method above.
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 the post on factoring) since this strategy is guaranteed to work (whereas sometimes factoring works well and sometimes not).
Here is the formula:
Suppose we have a quadratic equation of the form: Then, the solutions are:
Note that (provided ), 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:
This is a somewhat ugly equation so before proceeding, we do some multiplication to get rid of the fractions. Multiplying by 7, we get:
And before using our formula, it is crucial that all the terms are on one side, and the other side is equal to . Thus, we subtract to get:
Now, we can apply the formula to get:
= \frac{11 \pm 29}{18}
Thus, .
Using the quadratic formula, then, is fairly 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 “Frere Jacques,” you could sing the following:
Negative b, negative b Brother Jacques, Brother Jacques Frere Jacques, frere Jacques
Plusminus square root, plusminus square root Are you sleeping? Are you sleeping? Dormezvous? Dormezvous?
bsquared minus 4ac, bsquared minus 4ac Ring the matins! Ring the matins! Sonnez les matines! Sonnez les matines!
All over 2a, all over 2a Ding dang dong, ding dang dong Ding dang dong, ding dang dong
Practice Problems
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:
where is some constant (for ) and is a variable. Basically, polynomials are combinations of constants and the positive powers of multiplied by any coefficient.
We say that the highest exponent of a variable in a polynomial is its degree. So for example,
since if we simplify the last equation, the term goes away and we just get
8  8 = 0 \textrm{ has degree 0}
x^2 + x = 0
3x^2  2x + 5 = 0
\frac{x^2}{9}  x + 3 = x^2  2x + 8
\frac{17x^2}{9} + 3x  5 = 0
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:
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.
Method of Elimination
 Pick a variable we want to eliminate.
Let’s pick .
 Manipulate the equations so that the chosen variable has the same coefficient in all of them.
Now, if we look at our equations, we see that equation 4 does not have any ’s in it. So no matter what constant we multiply the equation by, we will not get a term with 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 , so no further manipulation of the equation is needed.
Now, we get:
And we still have our equation 4:
 Subtract the equations from one another, thereby eliminating our variable.
Note that this only applies to the equations that have in them. And remember to use all of the equations that include (but we only need to use each equation once).
Then, we get:
 Now, we should have four equations with at most four different variables. Pick another variable to eliminate.
Let’s eliminate .
 Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
 Subtract the equations from one another, thereby eliminating our variable.
 Now, we should have three equations with at most three different variables.
And we do.
 We can now solve this problem with the steps detailed in a previous post (see here %TODO add hyperlink).
Method of Substitution
 Pick a variable we want to substitute for.
Let’s choose .
 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:
 Substitute into the other equations.
Since we used the second equation to get our equation for , we now substitute for in the first, third, fourth, and fifth equations. We then get (after simplifying):
 Now, we should have four equations with at most four different variables.
And this is true.
 Pick another variable to substitute for.
Let’s pick .
 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
 Substitute into the other equations.
Since we used the first equation, we need to plug in our formula for into the second, third, and fourth equations. We then get (after simplifying):
 Now, we should have three equations with at most three different variables.
And we do.
 We can now solve this problem with the steps detailed in a previous post. (See %TODO add hyperlink)
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:
We can rewrite the first equation as:
Thereby giving us in terms of and some constants. We can now plug this into the second equation, giving us:
And simplifying, we get:
Now that we have the value of , we can plug it into one of our equations to get the value of :
In general, the steps of our method are as follows. Suppose we have some system of equations:
where are constants.
 Pick one variable to substitute.
For our example, let’s pick to substitute.
 Define that variable in terms of the other variables.
We can use either equation to define the value of . Let’s use the first. Then, we get:
 Substitute that variable into the remaining equation to solve for one of the variables.
Since we used the first equation to define , we need to plug it into the second equation. Then, we get:
 Now that you have one of the variables, use the equations to find the other.
Just plug in your value for into to find the value of .
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.
Suppose we have the following system of linear equations:
Now, we pick a variable that we want to substitute. Suppose we want to substitute first. The key is that we can pick any of the three equations to define in terms of the other variables, and . Then, we use that equation to substitute for in the other two equations.
Suppose we want to use equation 2 to define . Then, we get:
Then, since we used equation 2 to define , we need to substitute for in equations 1 and 3 (i.e. the other equations). We then get:
Simplifying, we get:
Which we can solve via our standard substitution methods.
Practice Problems
Solve the following systems of equations using the method of elimination





Solutions




In this post, we will talk about how to solve a system of linear equations via the method of elimination. Now, it is worth noting that you do not need to know both this method and the method of substitution that we will discuss next. You just need to be able to solve a system of linear equations. But for some problems, one or the other of these methods may be easier. So if you have the time, it is worth learning both.
Suppose we have a system of equations as follows:
Now, we could try to solve this system by guessing values of and , 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:
Or, in other words,
Now that we have found the value of , we can plug it back in to find a solution for :
So our solution is .
In general, the strategy for solving equations by elimination is to try and get certain 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 go back to one of the given equations
Let’s look at another example:
Here, it is not useful to simply subtract the second equation from the first. Doing so yields:
which is no more helpful than what we started with.
Instead, we need to make a decision: do we want to eliminate or ? It does not matter too much which one you choose, at most you can save yourself a few calculations. Let’s say we want to eliminate . Then, we need to manipulate our equations so that the variable will be eliminated. We see that in our second equation, has a coefficient of but in the first equation, only has a coefficient of .
[Graphic underlining the coefficients]
Subtracting these equations, then, will get us an equation that includes . To get the ’s to cancel out, we need to have the same coefficient in both equations. So let’s multiply the first equation by . (We could have also divided the second equation by 2, but fractions are harder to work with). Then, we get
And now, we can subtract the second equation from the first to get:
Plugging this back into one of our initial equations, we get:
So our solution is
In general, elimination problems will look like the following:
where are constants.
Here are some steps to solving these problems via elimination.
 Pick either or to eliminate.
For our example, suppose we want to eliminate .
 Force to have the same coefficient in both equations.
To do that here, we multiply the first equation by to get:
(The choice to manipulate the first equation was arbitrary. We could have instead multiplied the second equation by to get similar results.)
 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:
 Once we know the value of , we can just plug it in to get the value of .
Here, .
Now, we can also use the method of elimination to solve systems of 3 equations. Here, the key will be to turn our system of 3 equations into a system of 2 equations. This is done by picking a variable to eliminate. Then, we eliminate that variable using (roughly) the above method. Then, we apply our normal method for solving a system of 2 equations.
Here is an example.
Suppose we want to eliminate the variable . Then, we multiply our equations by constants so that will have the same coefficient in every equation. Since we want to avoid working with fractions, let’s try to make have the coefficient of 6 in every equation. Then, we multiply the first equation by , the second one by 2, and the third one by 3 to get:
Now, we need to cancel out the ’s. How do we do this? Now, as before, there are many ways to cancel out the ’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:
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





Solutions



