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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.

Respuesta :

MrRoyal

Answer:

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

Step-by-step explanation:

Given

[tex]3x - 3y = 1[/tex]

[tex]-2x+ 4y = 2[/tex]

Required

A good first step

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

Multiply [tex]3x - 3y = 1[/tex] by 2 (gotten from coefficient of x in the second equation)

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

This gives:

[tex]6x - 6y = 2[/tex]

[tex]-6x + 12y = 6[/tex]

Add both equations

[tex]6x - 6x - 6y + 12y = 2 + 6[/tex]

[tex]6y = 8[/tex]

[tex]y = \frac{4}{3}[/tex]

Hence (a), is true

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