Mathematics
archieforever
11

Solve Systems by Substitution x+y=-1 3x-2y=2 and x=y-2 3x-4y=-13

+1
(2) Answers
aideedeya

Problem one:  1.) Isolate one variable of one equation x+y-x=-1-x y=-1-x 2.) Plug in equation for isolated value into other original equation 3x-2(-1-x)=2 3.) Distribute 3x+2+2x=2 4.) Add like terms 5x+2=2 5.) Subtract two from both side 5x=0 6.) Divide both sides by 5 x=0 7.) Plug in numerical value for x into one equation (original or not) 0+y=1 8.) Subtract zero from both sides y=1 For the first problem, the answer is: x=0, y = 1 For the second problem, follow very similar steps: 3(y-2)-4y=-13 3y-6-4y=-13 -y-6=-13 -y=-7 y=7 x=y-2 x=7-2 x=5 That's your answer for the second problem: x=5, y=7 Hope this helps!

nikolak

1x + 1y = -1 ⇒ 3x - 3y = -3 3x -  2y = 2  ⇒ 3x - 2y = 2                                 -y = -5                                   y = 5                        3x - 2(5) = 2                          3x - 10 = 2                               +10  +10                                 3x = 12                                  3      3                                  x = 4                           (x , y) = (4, 5) 1x - 1y = -2   ⇒ 3x - 3y = -6 3x - 4y = -13 ⇒ 3x - 4y = -13                                    y = 7                         3x - 4(7) = -13                          3x - 28 = -13                               +28    +28                                 3x = 15                                  3      3                                   x = 5                             (x, y) = (5, 7)

Add answer