Question

How do you solve the system of equations `10x + 4y = 84` and `8x + 5y = 78`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for `40x`

• Equation 1:

`10x + 4y = 84`

`10x + 4y - color(red)(4y) = 84 - color(red)(4y)`

`10x + 0 = 84 - 4y`

`10x = 84 - 4y`

`color(red)(4) xx 10x = color(red)(4)(84 - 4y)`

`40x = (color(red)(4) xx 84) - (color(red)(4) xx 4y)`

`40x = 336 - 16y`

• Equation 2:

`8x + 5y = 78`

`8x + 5y - color(red)(5y) = 78 - color(red)(5y)`

`8x + 0 = 78 - 5y`

`8x = 78 - 5y`

`color(red)(5) xx 8x = color(red)(5)(78 - 5y)`

`40x = (color(red)(5) xx 78) - (color(red)(5) xx 5y)`

`40x = 390 - 25y`

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

`336 - 16y = 390 - 25y`

`336 - color(red)(336) - 16y + color(red)(25y) = 390 - color(red)(336) - 25y + color(red)(25y)`

`0 + (-16 + color(red)(25))y = 54 - 0`

`9y = 54`

`(9y)/color(red)(9) = 54/color(red)(9)`

`(color(red)(cancel(color(black)(9)))y)/cancel(color(red)(9)) = 6`

`y = 6`

Step 3) Substitute `6` for `y` in either of the equations in Step 1 and solve for `x`:

`40x = 390 - 25y` becomes:

`40x = 390 - (25 xx 6)`

`40x = 390 - 150`

`40x = 240`

`(40x)/color(red)(40) = 240/color(red)(40)`

`(color(red)(cancel(color(black)(40)))x)/cancel(color(red)(40)) = 6`

`x = 6`

The Solution Is:

`x = 6` and `y = 6`

Or

`(6, 6)`