Question

How do you solve the system `2p - 5q = 14` and `p + 3/2 q = 5`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve each equation for `2p`:

Equation 1

`2p - 5q = 14`

`2p - 5q + color(red)(5q) = 14 + color(red)(5q)`

`2p - 0 = 14 + 5q`

`2p = 14 + 5q`

Equation 2

`p + 3/2q = 5`

`p + 3/2q - color(red)(3/2q) = 5 - color(red)(3/2q)`

`p + 0 = 5 - 3/2q`

`p = 5 - 3/2q`

`color(red)(2) xx p = color(red)(2)(5 - 3/2q)`

`2p = (color(red)(2) xx 5) - (color(red)(2) xx 3/2q)`

`2p = 10 - 3q`

Step 2) Substitute `10 - 3q` from the second equation for `2p` in the first equation and solve for `p`:

`2p = 14 + 5q` becomes:

`10 - 3q = 14 + 5q`

`-color(blue)(14) + 10 - 3q + color(red)(3q) = -color(blue)(14) + 14 + 5q + color(red)(3q)`

`-4 - 0 = 0 + (5 + color(red)(3))q`

`-4 = 8q`

`-4/color(red)(8) = (8q)/color(red)(8)`

`-1/2 = (color(red)(cancel(color(black)(8)))q)/cancel(color(red)(8))`

`-1/2 = q` or `q = -1/2`

Step 3) Substitute `-1/2` for `q` in either of the equations from Step 1 and calculate `p`:

`2p = 14 + 5q` becomes:

`2p = 14 + (5 * -1/2)`

`2p = 14 - 5/2`

`2p = (2/2 * 14) - 5/2`

`2p = 28/2 - 5/2`

`2p = 23/2`

`color(red)(1/2) * 2p = color(red)(1/2) * 23/2`

`1p = 23/4`

`p = 23/4`

The Solution Is: `p = 23/4` and `q = -1/2`