Precalculus Examples

$x - z = 6$ , $y + 2 z = 3$

Step 1

Solve the equation for $y$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$y + 2 \cdot - 6 + 2 x = 3$

$z = - 6 + x$

Step 1.1.2

Multiply $2$ by $- 6$ .

$y - 12 + 2 x = 3$

$z = - 6 + x$

$y - 12 + 2 x = 3$

$z = - 6 + x$

Step 1.2

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

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Step 1.2.1

Add $12$ to both sides of the equation.

$y + 2 x = 3 + 12$

$z = - 6 + x$

Step 1.2.2

Subtract $2 x$ from both sides of the equation.

$y = 3 + 12 - 2 x$

$z = - 6 + x$

Step 1.2.3

Add $3$ and $12$ .

$y = 15 - 2 x$

$z = - 6 + x$

$y = 15 - 2 x$

$z = - 6 + x$

$y = 15 - 2 x$

$z = - 6 + x$

Step 2

Simplify the right side.

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Step 2.1

Reorder $15$ and $- 2 x$ .

$y = - 2 x + 15$

$z = - 6 + x$

$y = - 2 x + 15$

$z = - 6 + x$

Step 3

Simplify the right side.

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Step 3.1

Reorder $- 6$ and $x$ .

$z = x - 6$

$y = - 2 x + 15$

$z = x - 6$

$y = - 2 x + 15$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$