Solve by Substitution 6x+3y=33 , 4x+y=15

$6 x + 3 y = 33$ , $4 x + y = 15$

Step 1

Subtract $4 x$ from both sides of the equation.

$y = 15 - 4 x$

$6 x + 3 y = 33$

Step 2

Replace all occurrences of $y$ with $15 - 4 x$ in each equation.

Replace all occurrences of $y$ in $6 x + 3 y = 33$ with $15 - 4 x$ .

$6 x + 3 (15 - 4 x) = 33$

$y = 15 - 4 x$

Simplify the left side.

Simplify $6 x + 3 (15 - 4 x)$ .

Simplify each term.

Apply the distributive property.

$6 x + 3 \cdot 15 + 3 (- 4 x) = 33$

$y = 15 - 4 x$

Multiply $3$ by $15$ .

$6 x + 45 + 3 (- 4 x) = 33$

$y = 15 - 4 x$

Multiply $- 4$ by $3$ .

$6 x + 45 - 12 x = 33$

$y = 15 - 4 x$

$6 x + 45 - 12 x = 33$

$y = 15 - 4 x$

Subtract $12 x$ from $6 x$ .

$- 6 x + 45 = 33$

$y = 15 - 4 x$

$- 6 x + 45 = 33$

$y = 15 - 4 x$

$- 6 x + 45 = 33$

$y = 15 - 4 x$

$- 6 x + 45 = 33$

$y = 15 - 4 x$

Step 3

Solve for $x$ in $- 6 x + 45 = 33$ .

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

Subtract $45$ from both sides of the equation.

$- 6 x = 33 - 45$

$y = 15 - 4 x$

Subtract $45$ from $33$ .

$- 6 x = - 12$

$y = 15 - 4 x$

$- 6 x = - 12$

$y = 15 - 4 x$

Divide each term in $- 6 x = - 12$ by $- 6$ and simplify.

Divide each term in $- 6 x = - 12$ by $- 6$ .

$\frac{- 6 x}{- 6} = \frac{- 12}{- 6}$

$y = 15 - 4 x$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 x}{- 6} = \frac{- 12}{- 6}$

$y = 15 - 4 x$

Divide $x$ by $1$ .

$x = \frac{- 12}{- 6}$

$y = 15 - 4 x$

$x = \frac{- 12}{- 6}$

$y = 15 - 4 x$

$x = \frac{- 12}{- 6}$

$y = 15 - 4 x$

Simplify the right side.

Divide $- 12$ by $- 6$ .

$x = 2$

$y = 15 - 4 x$

$x = 2$

$y = 15 - 4 x$

$x = 2$

$y = 15 - 4 x$

$x = 2$

$y = 15 - 4 x$

Step 4

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = 15 - 4 x$ with $2$ .

$y = 15 - 4 \cdot 2$

$x = 2$

Simplify the right side.

Simplify $15 - 4 \cdot 2$ .

Multiply $- 4$ by $2$ .

$y = 15 - 8$

$x = 2$

Subtract $8$ from $15$ .

$y = 7$

$x = 2$

$y = 7$

$x = 2$

$y = 7$

$x = 2$

$y = 7$

$x = 2$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(2, 7)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 7)$

Equation Form:

$x = 2, y = 7$

Step 7

Solve by Substitution 6x+3y=33 , 4x+y=15

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