How do you solve the following linear system: y = 4x - 13 , 31x - 6y = 29 y=4x13,31x6y=29?

1 Answer
Jul 19, 2017

See a solution process below:

Explanation:

Step 1) Because the first equation is already solved for yy we can substitute (4x - 13)(4x13) for yy in the second equation and solve for xx:

31x - 6y = 2931x6y=29 becomes:

31x - 6(4x - 13) = 2931x6(4x13)=29

31x - (6 * 4x) + (6 * 13) = 2931x(64x)+(613)=29

31x - 24x + 78 = 2931x24x+78=29

(31 - 24)x + 78 = 29(3124)x+78=29

7x + 78 = 297x+78=29

7x + 78 - color(red)(78) = 29 - color(red)(78)7x+7878=2978

7x - 0 = -497x0=49

7x = -497x=49

(7x)/color(red)(7) = -49/color(red)(7)7x7=497

(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = -7

x = -7

Step 2) Substitute -7 for x in the first equation and calculate y:

y = 4x - 13 becomes:

y = (4 xx -7) - 13

y = -28 - 13

y = -41

The Solution Is: x = -7 and y = -41 or (-7, -41)