Question

How do you solve the system of equations `1/2x + 3y = 5/6` and `1/3x - 5y = 16/9`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve both equations for `x`:

• Equation 1:

`1/2x + 3y = 5/6`

`color(red)(2)(1/2x + 3y) = color(red)(2) xx 5/6`

`(color(red)(2) xx 1/2x) + (color(red)(2) xx 3y) = cancel(color(red)(2)) xx 5/(color(red)(cancel(color(black)(6)))3)`

`color(red)(2)/2x + 6y = 5/3`

`1x + 6y = 5/3`

`x + 6y = 5/3`

`x + 6y - color(red)(6y) = 5/3 - color(red)(6y)`

`x + 0 = 5/3 - 6y`

`x = 5/3 - 6y`

• Equation 2:

`1/3x - 5y = 16/9`

`color(red)(3)(1/3x - 5y) = color(red)(3) xx 16/9`

`(color(red)(3) xx 1/3x) - (color(red)(3) xx 5y) = cancel(color(red)(3)) xx 16/(color(red)(cancel(color(black)(9)))3)`

`color(red)(3)/3x - 15y = 16/3`

`1x - 15y = 16/3`

`x - 15y = 16/3`

`x - 15y + color(red)(15y) = 16/3 + color(red)(15y)`

`x - 0 = 16/3 + 15y`

`x = 16/3 + 15y`

Step 2) Because the left side of both equations are equal we can equate the right side of both equations and solve for `y`:

`5/3 - 6y = 16/3 + 15y`

`5/3 - color(blue)(16/3) - 6y + color(red)(6y) = 16/3 - color(blue)(16/3) + 15y + color(red)(6y)`

`(5 - color(blue)(16))/3 - 0 = 0 + (15 + color(red)(6))y`

`-11/3 = 21y`

`1/color(red)(21) xx -11/3 = (21y)/color(red)(21)`

`-(1 xx 11)/(color(red)(21) xx 3) = (color(red)(cancel(color(black)(21)))y)/cancel(color(red)(21))`

`-11/63 = y`

`y = -11/63`

Step 3) Substitute `-11/63` for `y` in the solution to either equation in Step 1 and calculate `x`:

`x = 16/3 + 15y` becomes:

`x = 16/3 + (15 xx -11/63)`

`x = (21/21 xx 16/3) + (15 xx -11)/63`

`x = 336/63 + (-165)/63`

`x = (336- 165)/63`

`x = 171/63`

The Solution Is:

`x = 171/63` and `y = -11/63`

Or

`(171/63, -11/63)`