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Solve the system of equations using elimination.15q – 4r = 625q + 8r = 86A.q = –7, r = –6B.q = 6, r = 7C.q = –7, r = –7

Sagot :

To solve the system we multiply the first equation by 2, then we get:

[tex]\begin{gathered} 30q-8r=124 \\ 5q+8t=86 \end{gathered}[/tex]

Now we add the equations:

[tex]\begin{gathered} 35q=210 \\ q=\frac{210}{35} \\ q=6 \end{gathered}[/tex]

Once we know the value of q we plug it in the first equation and solve for r:

[tex]\begin{gathered} 15(6)-4r=62 \\ 90-4r=62 \\ 28=4r \\ r=\frac{28}{4} \\ r=7 \end{gathered}[/tex]

Then q=6 and r=7 and the answer is B.

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