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# ACT Math Skill Review: Solving Equations

This is one of the most important components of the exam. Algebraic expressions will appear in many different forms, ranging from simple linear equations to complex word problems. Some questions may ask you to solve for a specific value of x, while others may ask you to solve for x in terms of another variable(s).

To prepare you for the exam, we will first review solving basic linear equations and equations involving radical expressions.

#### Solving for One Variable in Terms of Another

These types of questions are meant to test your understanding and use of the basic rules of arithmetic with regards to algebraic expressions. Let’s take a look at an example: This question is asking you to isolate x. This means getting x on one side of the equation and everything else on the other. Remember, whatever you do to one side of the equation, you have to do on the other.

First, subtract 3y from both sides to get: Then, divide both sides by 2 to get the value of x in terms of and z: #### Solving Quadratic Equations by Factoring

Don’t be surprised if you are asked to solve quadratic equations that can be factored. You will not be asked to use the quadratic formula to solve these equations. Let’s take a look at the following example: First, you’ll want to get the equation into standard quadratic format (equal to zero). To do this, subtract 7x from both sides: Next, factor the equation. You know you’ll have to have (x-?)(x-?) since the middle term is negative (-7x) and the final term is positive (12). The factors of 12 that add up to 7 are 3 and 4, leaving you with: Therefore, either (x - 4) = 0 or (x - 3) = 0. The values x = 4 and x = 3 satisfy the original equation.

#### Examples 1. The correct answer is C. First, you have to determine what the common variable is between the two equations. Both contain s. Since you want r in terms of t, you’ll have to first put s in terms of r and then do a bit of substitution. So first, take 2r = 5s. Divide both sides by 5. You get: 2r ⁄ 5 = s.
Next, substitute this new value for s into the second equation to get: 5(2r ⁄ 5) = 6t. The fives cancel out and you are left with: 2r = 6t. Divide both sides by 2 and you have the answer: r = 3t
1. The correct answer is G. Let’s get this equation into a more workable form by combining the factored items on the right of the equals sign to get: 2x + 6 = x2 + 8x + 15. Next, we’ll get all of our terms on one side: x2 + 6x + 9 = 0. Now you need to factor this equation and find the possible values for x. You end up with (x + 3)(x + 3) = 0. Thus, x must equal -3. If you aren’t paying attention, you might be tempted to choose choice F. However, the question isn’t asking for a possible value of x, but rather how many different solutions there are for x. In this case, there is just one.

2. The correct answer is B. Once again, the question isn’t in the most useful format. You’ll need to factor the equation that appears in the numerator. It factors to (x + 3)(x + 4). Since you have x + 4 in the denominator, those two values cancel each other out. Thus, you’re left with x + 3 = 5. Solve for x by subtracting 3 from each side and you are left with x = 2. Thank you for visiting MyCollegeOptions.org

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