How do you solve $3x+ 4y=27$ and $-8x+ y=33$ using substitution?

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How do you solve $3x+ 4y=27$ and $-8x+ y=33$ using substitution?

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Answer 1 · 2018-04-06T03:55:10

The solution is $(-3,9)$ .

Explanation:

Solve the system:

$"Equation 1":$ $3x+4y=27$

$"Equation 2":$ $-8x+y=33$

Both equations are linear equations in standard form. The solution to the system is the point that the two lines have in common. The $x$ and $y$ coordinates will be determined by substitution.

Solve Equation 1 for $x$ .

$3x+4y=27$

$3x=27-4y$

Divide both sides by $3$ .

$x=27/3-(4y)/3$

Simplify.

$x=9-(4y)/3$

Substitute $9-4y/3$ for $x$ in Equation 2. Solve for $y$ .

$-8x+y=33$

$-8(9-(4y)/3)+y=33$

$-72+(32y)/3+y=33$

Multiply $y$ by $3/3$ to create an equivalent fraction with $3$ in the denominator.

$-72+(32y)/3+yxx3/3=33$

$-72+(32y)/3+(3y)/3=33$

$-72+(35y)/3=33$

Add $72$ to both sides.

$(35y)/3=33+72$

Simplify.

$(35y)/3=105$

Multiply both sides by $3$ .

$35y=105xx3$

$35y=315$

Divide both sides by $35$

$y=315/35$

$y=9$

Substitute $9$ for $y$ in Equation 1. Solve for $x$ .

$3x+4(9)=27$

$3x+36=27$

$3x=27-36$

$3x=-9$

$x=(-9)/3$

$x=-3$

The solution is the point of intersection for the two lines, which is $(-3,9)$ .

graph{(3x+4y-27)(y-8x-33)=0 [-14.24, 11.07, 1.37, 14.03]}

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How do you solve $3x+ 4y=27$ and $-8x+ y=33$ using substitution?

1 Answer

Explanation:

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How do you solve 3x+ 4y=273x+ 4y=27 and -8x+ y=33-8x+ y=33 using substitution?

1 Answer

Explanation:

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How do you solve $3x+ 4y=27$ and $-8x+ y=33$ using substitution?