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Ainat [17]
3 years ago
7

Solve the system of equations by substitution.

Mathematics
2 answers:
ira [324]3 years ago
6 0

Answer:

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=3,\:y=-1

Step-by-step explanation:

3x - 4y = 13

5x + 4y = 11

isolate x for 3x-4y=13

\mathrm{Subsititute\:}x=\frac{13+4y}{3}

\begin{bmatrix}5\cdot \frac{13+4y}{3}+4y=11\end{bmatrix}

\frac{65+32y}{3}=11

now isolate y for \frac{65+32y}{3}=11

\frac{65+32y}{3}=11

65+32y=33

32y=-32

Divide both sides by 32

\frac{32y}{32}=\frac{-32}{32}

y=-1

\mathrm{For\:}x=\frac{13+4y}{3}

\mathrm{Subsititute\:}y=-1

x=\frac{13+4\left(-1\right)}{3}

  =\frac{13-4\cdot \:1}{3}

  =\frac{9}{3}

\mathrm{Divide\:the\:numbers:}\:\frac{9}{3}=3

  =3

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=3,\:y=-1

Angelina_Jolie [31]3 years ago
5 0
You can see how I get the answer in the picture below. Hope it can help you.

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