**Example Algebra Problem #1**

This problem presents you with a system of equations. As you can see, you have two variables, x and y, and you have two equations that use x and y. When you have multiple variables, the rule of thumb is that you need the same number of equations as you have variables to solve for them, and here, you have two variables and two equations, so you’re fine.

We’ll solve this system of equations using the substitution method.

In this method, you solve for one of the two variables in one equation, and then you plug it into the second equation.

Let’s solve for x in the first equation.

7x – 4y = -1 [add 4y to both sides]

7x = -1 + 4y [divide both sides by 7]

x =^{-1}⁄_{7}+^{4}⁄_{7}y

Now that we’ve solved for x, let’s plug it into the second equation

5x + 3y = 52 [plug in x from above]

5(^{-1}⁄_{7}+^{4}⁄_{7}y) + 3y = 52

See, now we only have one variable! Now we can solve for y.

5(

^{-1}⁄_{7}+^{4}⁄_{7}y) + 3y = 52 [distribute the 5 ]

^{-5}⁄_{7}+^{20}⁄_{7}y + 3y = 52 [multiply both sides by 7]

-5 + 20y + 21y = 364 [combine like terms]

-5 + 41y = 364 [add 5 to both sides]

41y = 369 [divide both sides by 41]

y = 9

Now that we have a number for y, we can plug it back into the equation for x, which we got earlier, to solve for x.

x =

^{-1}⁄_{7}+^{4}⁄_{7}y [plug in y]

x =^{-1}⁄_{7}+^{4}⁄_{7}(9) [simplify]

x =^{-1}⁄_{7}+^{36}⁄_{7 }[subtract]

x =^{35}⁄_{7}= 5

Now we have both x and y (boy, they turned out pretty). The question is asking us what the value of x – y is, so let’s just subtract them!

x – y = 5 – 9 = -4

So the answer is (A).

**Example Algebra Problem #2**

An algebra word problem! How fun. The goal with most word problems is to translate the sentences into algebra equations and then solve them.

Let’s start by tackling the actual question itself. The question asks, “what fraction of those attending are women?”

Since we don’t know how many women or men are at the dance, let’s use variables. Let W be the number of women at the dance, and let M be the number of men.

The fraction of the attendees at the dance that are women is “the number of women at the dance divided by all the people at the dance.” In algebraic form, this is written as

W / (W+M)

Now that we have that, let’s see how the problem gives us other information to help solve the problem. We’re told that ⅔ of the number of women at the dance is equal to ½ the number of men. In algebraic form, this is written as

^{2}⁄_{3}W =^{1}⁄_{2}M

Well we can’t solve for either variable (we need two equations to solve for two variables, and we only have one equation), so this means we don’t need actual numbers; we just need proportions. Let’s solve for M by multiplying both sides by 2.

^{4}⁄_{3}W = M

Now, let’s plug it back into our fraction up there.

W / (W+M) [plug in M]

W / (W +^{4}⁄_{3}W) [simplify denominator]

W / (^{7}⁄_{3}W) [simplify fraction]

^{3}⁄_{7}

As you can see, the fraction of the women is ^{3}⁄_{7}, which is answer choice (B).

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