Solving Systems of Equations by Addition

Line up the two equations vertically so that the x term, the y term and the constant line up. Then add to eliminate one variable. (If one variable doesn't eliminate multiply the entire equation by a number that will make one variable eliminate.) Solve the equation for the remaining variable. Substitute the answer back into one equation and solve for the other variable.

Examples:

 

x + 3y = 11 x - 3y = 5 2x = 16 x = 8	2x + 5y = 5 x + 2y = 1 2x + 5y = 5 -2x -4y = -2 (Equation 2 after being multiplied by -2) y = 3
Substitute into original equation.
8 + 3y = 11 -8 -8 3y = 3 y = 1	x + 2(3) = 1 x + 6 = 1 x = - 5

 

 

 

x + y = 4 x – y = 6  a. x = 5, y = 1  b. x = -5, y = -1  c. x = -5, y = 1  d. x = 5, y = -1  e. x = 6, y = 4	2x – 3y = -12 x – 3y = 3  a. x = 15, y = 6  b. x = -15, y = -6  c. x = 15, y = -6  d. x = -15, y = 6  e. x = -6, y = -15
27 – 5x – 3y =0 14 – 2x – 2y = 0  a. x = -3, y = -4  b. x = 3, y = 4  c. x = 4, y = 3  d. x = 3, y = -4  e. x = -3, y = 4	2x + y = 17 2x – y = -5  a. x = -3, y = 11  b. x = -3, y = -11  c. x = 3, y = 11  d. x = 11, y = 3  e. x = -11, y = -3
6x + 5y = 2 4x – 2y = 12  a. x = 2, y = -2  b. x = -2, y = -2  c. x = 4, y = -2  d. x = 4, y = -4  e. x = 2, y = -4	40 – 2x – 4y =0 40 – 3x -2y =0  a. x = 5, y = -5  b. x = -10, y = -5  c. x = -10, y = 5  d. x = 10, y = -5  e. x = 10, y = 5

 

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