Algebra Examples

$z = 3$ , $x = 3 - 5 z$

Step 1

Solve the system of equations.

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Step 1.1

Simplify the right side.

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Step 1.1.1

Reorder $3$ and $- 5 z$ .

$x = - 5 z + 3$

$z = 3$

$x = - 5 z + 3$

$z = 3$

Step 1.2

Add $5 z$ to both sides of the equation.

$z = 3, x + 5 z = 3$

Step 1.3

Multiply each equation by the value that makes the coefficients of $z$ opposite.

$(- 5) \cdot (z) = (- 5) (3)$

$x + 5 z = 3$

Step 1.4

Simplify.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

Multiply $- 5$ by $z$ .

$- 5 z = (- 5) (3)$

$x + 5 z = 3$

$- 5 z = (- 5) (3)$

$x + 5 z = 3$

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Multiply $- 5$ by $3$ .

$- 5 z = - 15$

$x + 5 z = 3$

$- 5 z = - 15$

$x + 5 z = 3$

$- 5 z = - 15$

$x + 5 z = 3$

Step 1.5

Add the two equations together to eliminate $z$ from the system.

	$-$	$5$	$z$	$=$	$-$	$1$	$5$
$+$	$+$	$5$	$z$	$=$			$3$
$x$				$=$	$-$	$1$	$2$

Step 1.6

Substitute the value found for $x$ into one of the original equations, then solve for $z$ .

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Step 1.6.1

Substitute the value found for $x$ into one of the original equations to solve for $z$ .

$(- 12) + 5 z = 3$

Step 1.6.2

Remove parentheses.

$- 12 + 5 z = 3$

Step 1.6.3

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

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Step 1.6.3.1

Add $12$ to both sides of the equation.

$5 z = 3 + 12$

Step 1.6.3.2

Add $3$ and $12$ .

$5 z = 15$

Step 1.6.4

Divide each term in $5 z = 15$ by $5$ and simplify.

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Step 1.6.4.1

Divide each term in $5 z = 15$ by $5$ .

$\frac{5 z}{5} = \frac{15}{5}$

Step 1.6.4.2

Simplify the left side.

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Step 1.6.4.2.1

Cancel the common factor of $5$ .

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Step 1.6.4.2.1.1

Cancel the common factor.

$\frac{5 z}{5} = \frac{15}{5}$

Step 1.6.4.2.1.2

Divide $z$ by $1$ .

$z = \frac{15}{5}$

Step 1.6.4.3

Simplify the right side.

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Step 1.6.4.3.1

Divide $15$ by $5$ .

$z = 3$

Step 1.7

This is the final solution to the independent system of equations.

$x = - 12$

$z = 3$

$x = - 12$

$z = 3$

Step 2

Since the system has a point of intersection, the system is independent.

Independent