Question

When solving the system of equations 3x−3y=1 and −2x+4y=2 algebraically, a good first step is to:

A. Multiply 3x−3y=1 by 2, and multiply −2x+4y=2 by 3.
B. Multiply 3x−3y=1 by 3, and multiply −2x+4y=2 by −2.
C. Rewrite both equations in slope-intercept form.
D. Add the equations together.

Answer 1

A. Multiply 3x−3y=1 by 2, and multiply −2x+4y=2 by 3.

Explanation:

Given

`3x - 3y = 1`

`-2x+ 4y = 2`

Required

A good first step

A good first step is to eliminate x and this is done as follows:

Multiply
`3x - 3y = 1` by 2 (gotten from coefficient of x in the second equation)

Multiply
`-2x+ 4y = 2` by 3 (gotten from coefficient of x in the first equation)

This gives:

`6x - 6y = 2`

`-6x + 12y = 6`

Add both equations

`6x - 6x - 6y + 12y = 2 + 6`

`6y = 8`

`y = (4)/(3)`

Hence (a), is true