what’s the solution to x-4y=-8 and 2x-3y=-16

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carlosego carlosego · Answer 1 · 2019-03-29T21:16:13+01:00

Answer: [tex]x=-8\\y=0[/tex]

Step-by-step explanation:

You can apply the elimination method:

- Multiply the first equation by -2.

- Add both equations to cancel out the variable x.

- Solve for y:

[tex]\left \{ {{(-2)(x-4y)=-8(-2)} \atop {2x-3y=-16}} \right.\\\\\\\left \{ {{-2x+8y=16} \atop {2x-3y=-16}} \right.\\ ------\\5y=0\\y=0[/tex]

- Substitute y=0 into any of the original equations ans solve for x. Then:

[tex]x-4(0)=-8\\x=-8[/tex]

mixter17 mixter17 · Answer 2 · 2019-03-29T21:41:03+01:00

The answers are:

[tex]x=-8\\y=0[/tex]

Since we have to equations, first we need to isolate one variable, and then substitute it into the other equation.

So, isolating x from the first equation we have:

[tex]x-4y=-8\\x=-8+4y[/tex]

Then, substituting "x" into the second equation, we have:

[tex]2(-8+4y)-3y=-16\\-16+8y-3y-=-16\\5y=-16+16\\5y=0\\y=\frac{0}{5}=0[/tex]

Therefore, substituting "y" into the first equation, we have:

[tex]x-4(0)=-8\\x-0=-8\\x=-8[/tex]

So, the solutions for the system of equation are:

[tex]x=-8\\y=0[/tex]

Have a nice day!