Solve by Addition/Elimination x+2y+z=4 , 4y-3z=1 , 2y+10z=12

$x + 2 y + z = 4$ , $4 y - 3 z = 1$ , $2 y + 10 z = 12$

Step 1

Choose two equations and eliminate one variable. In this case, eliminate $z$ .

$x + 2 y + z = 4$

$4 y - 3 z = 1$

Step 2

Eliminate $z$ from the system.

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

$(3) \cdot (x + 2 y + z) = (3) (4)$

$4 y - 3 z = 1$

Simplify.

Simplify the left side.

Simplify $(3) \cdot (x + 2 y + z)$ .

Apply the distributive property.

$3 x + 3 (2 y) + 3 z = (3) (4)$

$4 y - 3 z = 1$

Multiply $2$ by $3$ .

$3 x + 6 y + 3 z = (3) (4)$

$4 y - 3 z = 1$

$3 x + 6 y + 3 z = (3) (4)$

$4 y - 3 z = 1$

$3 x + 6 y + 3 z = (3) (4)$

$4 y - 3 z = 1$

Simplify the right side.

Multiply $3$ by $4$ .

$3 x + 6 y + 3 z = 12$

$4 y - 3 z = 1$

$3 x + 6 y + 3 z = 12$

$4 y - 3 z = 1$

$3 x + 6 y + 3 z = 12$

$4 y - 3 z = 1$

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

			$3$	$x$	$+$	$6$	$y$		$+$	$3$	$z$	$=$	$1$	$2$
$+$							$4$	$y$	$-$	$3$	$z$	$=$		$1$
			$3$	$x$	$+$	$1$	$0$	$y$				$=$	$1$	$3$

The resultant equation has $z$ eliminated.

$3 x + 10 y = 13$

$3 x + 10 y = 13$

Step 3

Choose another two equations and eliminate $z$ .

$4 y - 3 z = 1$

$2 y + 10 z = 12$

Step 4

Eliminate $z$ from the system.

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

$(10) \cdot (4 y - 3 z) = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

Simplify.

Simplify the left side.

Simplify $(10) \cdot (4 y - 3 z)$ .

Apply the distributive property.

$10 (4 y) + 10 (- 3 z) = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

Multiply.

Multiply $4$ by $10$ .

$40 y + 10 (- 3 z) = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

Multiply $- 3$ by $10$ .

$40 y - 30 z = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

$40 y - 30 z = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

$40 y - 30 z = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

$40 y - 30 z = (10) (1)$

$(3) \cdot (2 y + 10 z) = (3) (12)$

Simplify the right side.

Multiply $10$ by $1$ .

$40 y - 30 z = 10$

$(3) \cdot (2 y + 10 z) = (3) (12)$

$40 y - 30 z = 10$

$(3) \cdot (2 y + 10 z) = (3) (12)$

Simplify the left side.

Simplify $(3) \cdot (2 y + 10 z)$ .

Apply the distributive property.

$40 y - 30 z = 10$

$3 (2 y) + 3 (10 z) = (3) (12)$

Multiply.

Multiply $2$ by $3$ .

$40 y - 30 z = 10$

$6 y + 3 (10 z) = (3) (12)$

Multiply $10$ by $3$ .

$40 y - 30 z = 10$

$6 y + 30 z = (3) (12)$

$40 y - 30 z = 10$

$6 y + 30 z = (3) (12)$

$40 y - 30 z = 10$

$6 y + 30 z = (3) (12)$

$40 y - 30 z = 10$

$6 y + 30 z = (3) (12)$

Simplify the right side.

Multiply $3$ by $12$ .

$40 y - 30 z = 10$

$6 y + 30 z = 36$

$40 y - 30 z = 10$

$6 y + 30 z = 36$

$40 y - 30 z = 10$

$6 y + 30 z = 36$

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

			$4$	$0$	$y$	$-$	$3$	$0$	$z$	$=$	$1$	$0$
$+$				$6$	$y$	$+$	$3$	$0$	$z$	$=$	$3$	$6$
			$4$	$6$	$y$					$=$	$4$	$6$

The resultant equation has $z$ eliminated.

$46 y = 46$

$46 y = 46$

Step 5

Take the resultant equations and eliminate another variable. In this case, eliminate $y$ .

$3 x + 10 y = 13$

$46 y = 46$

Step 6

Eliminate $y$ from the system.

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

$(- 23) \cdot (3 x + 10 y) = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

Simplify.

Simplify the left side.

Simplify $(- 23) \cdot (3 x + 10 y)$ .

Apply the distributive property.

$- 23 (3 x) - 23 (10 y) = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

Multiply.

Multiply $3$ by $- 23$ .

$- 69 x - 23 (10 y) = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

Multiply $10$ by $- 23$ .

$- 69 x - 230 y = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

$- 69 x - 230 y = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

$- 69 x - 230 y = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

$- 69 x - 230 y = (- 23) (13)$

$(5) \cdot (46 y) = (5) (46)$

Simplify the right side.

Multiply $- 23$ by $13$ .

$- 69 x - 230 y = - 299$

$(5) \cdot (46 y) = (5) (46)$

$- 69 x - 230 y = - 299$

$(5) \cdot (46 y) = (5) (46)$

Simplify the left side.

Multiply $46$ by $5$ .

$- 69 x - 230 y = - 299$

$230 y = (5) (46)$

$- 69 x - 230 y = - 299$

$230 y = (5) (46)$

Simplify the right side.

Multiply $5$ by $46$ .

$- 69 x - 230 y = - 299$

$230 y = 230$

$- 69 x - 230 y = - 299$

$230 y = 230$

$- 69 x - 230 y = - 299$

$230 y = 230$

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

			$-$	$6$	$9$	$x$	$-$	$2$	$3$	$0$	$y$	$=$	$-$	$2$	$9$	$9$
$+$								$2$	$3$	$0$	$y$	$=$		$2$	$3$	$0$
			$-$	$6$	$9$	$x$						$=$		$-$	$6$	$9$

The resultant equation has $y$ eliminated.

$- 69 x = - 69$

Divide each term in $- 69 x = - 69$ by $- 69$ and simplify.

Divide each term in $- 69 x = - 69$ by $- 69$ .

$\frac{- 69 x}{- 69} = \frac{- 69}{- 69}$

Simplify the left side.

Cancel the common factor of $- 69$ .

Cancel the common factor.

$\frac{- 69 x}{- 69} = \frac{- 69}{- 69}$

Divide $x$ by $1$ .

$x = \frac{- 69}{- 69}$

$x = \frac{- 69}{- 69}$

$x = \frac{- 69}{- 69}$

Simplify the right side.

Divide $- 69$ by $- 69$ .

$x = 1$

$x = 1$

$x = 1$

$x = 1$

Step 7

Substitute the value of $x$ into an equation with $z$ eliminated already and solve for the remaining variable.

Substitute the value of $x$ into an equation with $z$ eliminated already.

$3 (1) + 10 y = 13$

Solve for $y$ .

Multiply $3$ by $1$ .

$3 + 10 y = 13$

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

Subtract $3$ from both sides of the equation.

$10 y = 13 - 3$

Subtract $3$ from $13$ .

$10 y = 10$

$10 y = 10$

Divide each term in $10 y = 10$ by $10$ and simplify.

Divide each term in $10 y = 10$ by $10$ .

$\frac{10 y}{10} = \frac{10}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{10}{10}$

Divide $y$ by $1$ .

$y = \frac{10}{10}$

$y = \frac{10}{10}$

$y = \frac{10}{10}$

Simplify the right side.

Divide $10$ by $10$ .

$y = 1$

$y = 1$

$y = 1$

$y = 1$

$y = 1$

Step 8

Substitute the value of each known variable into one of the initial equations and solve for the last variable.

Substitute the value of each known variable into one of the initial equations.

$(1) + 2 (1) + z = 4$

Solve for $z$ .

Simplify $(1) + 2 (1) + z$ .

Multiply $2$ by $1$ .

$1 + 2 + z = 4$

Add $1$ and $2$ .

$3 + z = 4$

$3 + z = 4$

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

Subtract $3$ from both sides of the equation.

$z = 4 - 3$

Subtract $3$ from $4$ .

$z = 1$

$z = 1$

$z = 1$

$z = 1$

Step 9

The solution to the system of equations can be represented as a point.

$(1, 1, 1)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(1, 1, 1)$

Equation Form:

$x = 1, y = 1, z = 1$

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