How do you solve the following system: 2x+y=5, -29=5y-3x 2x+y=5,29=5y3x?

1 Answer
Jul 27, 2017

See a solution process below:

Explanation:

Step 1) Solve the first equation for yy:

2x + y = 52x+y=5

-color(red)(2x) + 2x + y = -color(red)(2x) + 52x+2x+y=2x+5

0 + y = -2x + 50+y=2x+5

y = -2x + 5y=2x+5

Step 2) Substitute (-2x + 5)(2x+5) for yy in the second equation and solve for xx:

-29 = 5y - 3x29=5y3x becomes:

-29 = 5(-2x + 5) - 3x29=5(2x+5)3x

-29 = (5 xx -2x) + (5 xx 5) - 3x29=(5×2x)+(5×5)3x

-29 = -10x + 25 - 3x29=10x+253x

-29 = -10x - 3x + 2529=10x3x+25

-29 = (-10 - 3)x + 2529=(103)x+25

-29 = -13x + 2529=13x+25

-29 - color(red)(25) = -13x + 25 - color(red)(25)2925=13x+2525

-54 = -13x + 054=13x+0

-54 = -13x54=13x

(-54)/color(red)(-13) = (-13x)/color(red)(-13)5413=13x13

54/13 = (color(red)(cancel(color(black)(-13)))x)/cancel(color(red)(-13))

54/13 = x

x = 54/13

Step 3) Substitute 54/13 for x in the solution to the first equation at the end of Step 1 and calculate y:

y = -2x + 5 becomes:

y = (-2 xx 54/13) + 5

y = -108/13 + 5

y = -108/13 + (13/13 xx 5)

y = -108/13 + 65/13

y = -43/13

The Solution Is: x = 54/13 and y = -43/13 or (54/13, -43/13)