Solve by Substitution y=9+3z , y=51-3z

$y = 9 + 3 z$ , $y = 51 - 3 z$

Eliminate the equal sides of each equation and combine.

$9 + 3 z = 51 - 3 z$

Solve $9 + 3 z = 51 - 3 z$ for $z$ .

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Move all terms containing $z$ to the left side of the equation.

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Add $3 z$ to both sides of the equation.

$9 + 3 z + 3 z = 51$

Add $3 z$ and $3 z$ .

$9 + 6 z = 51$

Move all terms not containing $z$ to the right side of the equation.

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Subtract $9$ from both sides of the equation.

$6 z = 51 - 9$

Subtract $9$ from $51$ .

$6 z = 42$

Divide each term by $6$ and simplify.

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Divide each term in $6 z = 42$ by $6$ .

$\frac{6 z}{6} = \frac{42}{6}$

Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 z}{6} = \frac{42}{6}$

Divide $z$ by $1$ .

$z = \frac{42}{6}$

Divide $42$ by $6$ .

$z = 7$

Evaluate $y$ when $z = 7$ .

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Substitute $7$ for $z$ .

$y = 51 - 3 \cdot 7$

Substitute $7$ for $z$ in $y = 51 - 3 \cdot 7$ and solve for $y$ .

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Remove parentheses.

$y = 51 - 3 \cdot 7$

Simplify $51 - 3 \cdot 7$ .

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Multiply $- 3$ by $7$ .

$y = 51 - 21$

Subtract $21$ from $51$ .

$y = 30$

The solution of the system of equations is all the values that make the system true.

$y = 30, z = 7$

Solve by Substitution y=9+3z , y=51-3z

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