8x - 3y = -3 5x - 2y = -1 These are two linear equations that you must solve at the same time, meaning you must use both equations to solve both equations.

In the case of our example, you are trying to find out what numbers ‘x’ and ‘y’ represent that will make both of the equations true. In the case of this example, x = -3 and y = -7. Plug them in. 8(-3) - 3(-7) = -3. This is TRUE. 5(-3) -2(-7) = -1. This is also TRUE.

8 and 3 for the first equation; 5 and 2 for the second equation.

Substitution, on the other hand, is where you begin working with only one equation so that you can again solve for one variable. Once you solve one equation, you can plug in your findings to the other equation, effectively making one large equation out of your two smaller ones. Again, don’t worry—this will be covered in detail in Method 3.

8x - 3y = -3 5x - 2y = -1

In 8x - 3y = -3 (equation A) and 5x - 2y = -1 (equation B), you can multiply equation A with 2 and equation B with 3 so that you get 6y in equation A and 6y in equation B. This would look like: equation A: 2(8x - 3y =-3) = 16x -6y = -6. Equation B: 3(5x - 2y = -1) = 15x -6y =-3

(16x - 6y = -6) - (15x - 6y = -3) = 1x = -3. Therefore x = -3. For other cases, if the numerical coefficient of x is not 1 after we add or subtract, we must divide both sides by the numerical coefficient to simplify the equation.

Equation B: 5(-3) - 2y = -1 so -15 -2y = -1. Add 15 to both sides so -2y = 14. Divide both sides by -2 so that y = -7. Therefore x = -3 and y = -7.

8(-3) - 3(-7) = -3 so -24 +21 = -3 TRUE. 5(-3) -2(-7) = -1 so -15 + 14 = -1 TRUE. Therefore, the variables we have found are correct.

x - 2y = 10 (equation A) and -3x -4y = 10 (equation B). You would choose to work with x - 2y = 10 because the coefficient of x in this equation is 1. Solving for x in equation A would meaning adding 2y to both sides. Therefore, x = 10 + 2y.

Insert the ‘x’ of equation B into equation A: -3(10 + 2y) -4y = 10. You can see that we have taken ‘x’ out of the equation and inserted what ‘x’ equals.

-3(10 + 2y) -4y = 10 so -30 -6y -4y = 10. Combine the y’s: -30 - 10y = 10. Move the -30 over to the other side: -10y = 40. Solve for y: y = -4.

Solve for ‘x’ in equation A by plugging in y = -4: x - 2(-4) = 10. Simply the equation: x + 8 = 10. Solve for x: x = 2.

Equation A: 2 - 2(-4) = 10 is TRUE. Equation B: -3(2) -4(-4) = 10 is TRUE.