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Test PrepSATMathLinear Equations

SAT Math Skill Review: Linear Equations

You will sometimes be asked to solve systems of two or more linear equations or inequalities. Linear systems are equations that contain the same variables. For example, a + b = 11 and 2a + b = 10 are considered linear systems since both contain the same variables, a and b.

To solve systems of linear equations, you can use the substitution method:

  • Take one of the equations listed and find the value of one of the variables in terms of the other.
  • Substitute the value found for the variable into the other equation.
  • Solve for the second variable.
  • Substitute that value into the original equation to solve for the first variable.

Let’s see how this works:

For what values of x and y are the following equations true?

In the first equation, solve for x in terms of y.

Then substitute this value for x into the second equation and solve for y.

Substitute y = 9 back into the original equation to find the value of x.

Examples

Try the following examples on your own:

Linear Equations Examples 

Answers and Explanations

  1. The correct answer is A. You want to solve for x, so let’s start by putting y in terms of x, using the second equation: y = 4 – 2x. Substitute that value for y back into the first equation: 3x + 5(4 – 2x) = 6. This becomes 3x + 20 – 10x = 6, and then -7x = -14. Thus, x = 2.
  2. The correct answer is D. If you have 3 variables, you need 3 equations. It’s the same process as with two equations, you just have a few extra steps. Start with the third equation and put x in terms of y and z. You end up with 2x = zy. Stick with 2x as it’ll be easier to work with when substituting it back into the other equations.  Take this value and substitute it into the second equation for x. You end up with 2(z - y) + y – 3z = -6. This becomes 2z – 2y + y – 3z = -6. From there, isolate y to get y in terms of z: -z - y = -6,or y = 6 - z. Now plug this value for y into the value for 2x, to get x in terms of z: 2x = z – (6 - z). This becomes 2x = 2z – 6. You now have values for both 2x and y in terms of z. Substitute these values into the first equation to get: 4(2z – 6) + 3(6 – z) + 2z = 22. Continue to simplify: 8z – 24 + 18 – 3z + 2z = 22 and 7z – 6 = 22. Finally, 7z = 28 and z = 4.   

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