SOLUTION: how do i solve system of equations? that look like these 1.) 5x-3y=-4 x+2y=7 i tried substitution : i moved the 5x to th

Algebra ->  Systems-of-equations -> SOLUTION: how do i solve system of equations? that look like these 1.) 5x-3y=-4 x+2y=7 i tried substitution : i moved the 5x to th      Log On


   

Question 692125: how do i solve system of equations? that look like these
1.) 5x-3y=-4
x+2y=7

i tried substitution : i moved the 5x to the other side and got ----> -3y=-4-5x
__-3y=-4-5x__ then substituted it in the second equation ---> x+2(-4/-3-5/-3x)=7 then i multiplied 2 to (-4/-3-5/-3x)
-3
then i got x-8/-3-10/-3x=7 and then i got confused and i didn't know what to do .










Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
5x-3y=-4
x+2y=7
...
Using substitution, solve for y
-3y = -4 - 5x
y = 4%2F3+%2B+%285%2F3%29x ==> the negative divided by a negative cancel out
then, substitute
x + 2(4%2F3%2B%285%2F3%29x = 7 ==> distribute the 2
x + 8%2F3+%2B+%2810%2F3%29x = 7 ==> multiply both sides by 3 to remove the fraction
3x + 8 + 10x = 21 ==> combine like terms
13x = 21 - 8
13x = 13
x = 1
...
Rearrange x+2y=7 to solve for x
x = 7-2y, then substitute
5(7-2y) - 3y = -4 ==> distribute the 5
35 - 10y - 3y = -4
-13y = - 4 - 35
-13y = -39
y = 3
...
Solution
(highlight_green%281%29,highlight_green%283%29)
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