How do you solve x² + y² = 20 and x + y = 6 using substitution?

1 Answer
Sep 5, 2017

See a solution process below:

Explanation:

Step 1) Solve the second equation for x:

x + y = 6

x + y - color(red)(y) = 6 - color(red)(y)

x + 0 = 6 - y

x = 6 - y

Step 2) Substitute (6 - y) for x in the first equation and solve for y:

x^2 + y^2 = 20 becomes:

(6 - y)^2 + y^2 = 20

36 - 12y + y^2 + y^2 = 20

1y^2 + 1y^2 - 12y + 36 = 20

(1 + 1)y^2 - 12y + 36 = 20

2y^2 - 12y + 36 = 20

2y^2 - 12y + 36 - color(red)(20) = 20 - color(red)(20)

2y^2 - 12y + 16 = 0

(2y - 8)(y - 2) = 0

Solution 1 for y:

2y - 8 = 0

2y - 8 + color(red)(8) = 0 + color(red)(8)

2y - 0 = 8

2y = 8

(2y)/color(red)(2) = 8/color(red)(2)

(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = 4

y = 4

Solution 2 for y:

y - 2 = 0

y - 2 + color(red)(2) = 0 + color(red)(2)

y - 0 = 2

y = 2

Step 3) Substitute the values for y into the solution to the second equation at the end of Step 1 and calculate x:

x = 6 - y and x = 6 - y becomes:

x = 6 - 4 and x = 6 - 2

x = 2 and x = 4

The Solutions Are:

x = 2 and y = 4 or (2, 4)

AND

x = 4 and y = 2 or (4, 2)