# Mathematics - Writing Equations

Select any two integers between -12 and +12 which will become solutions to a system of two equations.
Write two equations that have your two integers as solutions. Show how you built the equations using your integers. Your solution and equations should not be the same as those of other students or the textbook. There are infinite possibilities.
Solve your system of equations by the addition/subtraction method. Make sure you show the necessary 5 steps.

### Comments for Mathematics - Writing Equations

 Sep 14, 2011 Mathematics - Writing Equations by: Staff ---------------------------------------------------- Part II 5. Check your answer Check the answer. Solve the two equations using the addition/subtraction method. The solutions should be: x = -7, y = 2 x + 3y = -1 5x - y = -37 Multiply the 2nd equation by 3 3*(5x - y = -37) 3*5x - 3*y = 3*(-37) 15x - 3y = -111 The two equations are now: x + 3y = -1 15x - 3y = -111 Add the two equations to eliminate the y variable: x + 3y = -1 + (15x - 3y = -111) ------------------------ x + 15x + 3y + (- 3y) = -1 + (- 111) x + 15x + 3y - 3y = -1 – 111 Combine like terms (x + 15x) + (3y - 3y) = (-1 – 111) (16x) + (0) = (-112) 16x + 0 = -112 16x = -112 Divide each side of the equation by 16 16x/16 = -112/16 x*(16/16) = -112/16 x*(1) = -112/16 x*1 = -112/16 x = -112/16 x = -7 Substitute -7 for x in either of the two original equations, and then solve for y. x + 3y = -1 -7 + 3y = -1 Add 7 to each side of the equation: -7 + 3y + 7 = -1 + 7 Combine like terms: -7 + 7 + 3y = -1 + 7 (-7 + 7) + 3y = (-1 + 7) (0) + 3y = (6) 0 + 3y = 6 3y = 6 Divide each side of the equation by 3 3y/3 = 6/3 y*(3/3) = 6/3 y*(1) = 6/3 y*1 = 6/3 y = 6/3 y = 2 when the two equations are solved algebraically, the solution is: x = -7 y = 2 These values match the original integers selected at the beginning of the problem. Therefore, the equations are valid. Thanks for writing. Staff www.solving-math-problems.com