Question

How do you solve the system of equations `2p + 3q = 10` and `2q - 4p = 44`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for `4p`:

Equation 1:

`2p + 3q = 10`

`2p + 3q - color(red)(3q) = 10 - color(red)(3q)`

`2p + 0 = 10 - 3q`

`2p = 10 - 3q`

`color(red)(2) * 2p = color(red)(2)(10 - 3q)`

`4p = (color(red)(2) * 10) - (color(red)(2) * 3q)`

`4p = 20 - 6q`

Equation 2:

`2q - 4p = 44`

`-color(blue)(44) + 2q - 4p + color(red)(4p) = -color(blue)(44) + 44 + color(red)(4p)`

`-44 + 2q - 0 = 0 + 4p`

`-44 + 2q = 4p`

`4p = -44 + 2q`

Step 2) Because the left side of each equation is `4p` we can equate the right side of each equation and solve for `q`:

`20 - 6q = -44 + 2q`

`color(red)(44) + 20 - 6q + color(red)(6q) = color(red)(44) - 44 + 2q + color(red)(6q)`

`64 - 0 = 0 + (2 + color(red)(6))q`

`64 = 8q`

`64/color(red)(8) = (8q)/color(red)(8)`

`8 = (color(red)(cancel(color(black)(8)))q)/cancel(color(red)(8))`

`8 = q`

`q = 8`

*Step 3) Substitute `8` for `q` in the solution to either equation in Step 1 and calculate `p`:

`4p = 20 - 6q` becomes:

`4p = 20 - (6 * 8)`

`4p = 20 - 48`

`4p = -28`

`(4p)/color(red)(4) = -28/color(red)(4)`

`(color(red)(cancel(color(black)(4)))p)/cancel(color(red)(4)) = -7`

`p = -7`

The Solution Is: `p = -7` and `q = 8`