How do you solve the following linear system: x+y=4, -3x+y=-8 x+y=4,3x+y=8?

1 Answer
May 31, 2018

See a solution process below:

Explanation:

Step 1) Solve the first equation for xx:

x + y = 4x+y=4

x + y - color(red)(y) = 4 - color(red)(y)x+yy=4y

x + 0 = 4 - yx+0=4y

x = 4 - yx=4y

Step 2) Substitute (4 - y)(4y) for xx in the second equation and solve for yy:

-3x + y = -83x+y=8 becomes:

-3(4 - y) + y = -83(4y)+y=8

(-3 xx 4) + (3 xx y) + y = -8(3×4)+(3×y)+y=8

-12 + 3y + y = -812+3y+y=8

-12 + color(red)(12) + 3y + y = -8 + color(red)(12)12+12+3y+y=8+12

0 + 3y + y = 40+3y+y=4

3y + y = 43y+y=4

3y + 1y = 43y+1y=4

(3 + 1)y = 4(3+1)y=4

4y = 44y=4

(4y)/color(red)(4) = 4/color(red)(4)4y4=44

y = 1y=1

Step 3) Substitute 11 for yy in the solution to the first equation at the end of Step 1 and calculate xx:

x = 4 - yx=4y becomes:

x = 4 - 1x=41

x = 3x=3

The Solution Is:

x = 3x=3 and y = 1y=1

Or

(3, 1)(3,1)