Solve by Substitution 2x+y=7 , 5x+y=9

$2 x + y = 7$ , $5 x + y = 9$

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

$y = 7 - 2 x$

$5 x + y = 9$

Replace all occurrences of $y$ with $7 - 2 x$ in each equation.

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Replace all occurrences of $y$ in $5 x + y = 9$ with $7 - 2 x$ .

$5 x + 7 - 2 x = 9$

$y = 7 - 2 x$

Simplify $5 x + 7 - 2 x$ .

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Remove parentheses.

$5 x + 7 - 2 x = 9$

$y = 7 - 2 x$

Subtract $2 x$ from $5 x$ .

$3 x + 7 = 9$

$y = 7 - 2 x$

$3 x + 7 = 9$

$y = 7 - 2 x$

$3 x + 7 = 9$

$y = 7 - 2 x$

Solve for $x$ in the first equation.

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

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Subtract $7$ from both sides of the equation.

$3 x = 9 - 7$

$y = 7 - 2 x$

Subtract $7$ from $9$ .

$3 x = 2$

$y = 7 - 2 x$

$3 x = 2$

$y = 7 - 2 x$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

$y = 7 - 2 x$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

$y = 7 - 2 x$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

$y = 7 - 2 x$

$x = \frac{2}{3}$

$y = 7 - 2 x$

$x = \frac{2}{3}$

$y = 7 - 2 x$

$x = \frac{2}{3}$

$y = 7 - 2 x$

Replace all occurrences of $x$ with $\frac{2}{3}$ in each equation.

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Replace all occurrences of $x$ in $y = 7 - 2 x$ with $\frac{2}{3}$ .

$y = 7 - 2 (\frac{2}{3})$

$x = \frac{2}{3}$

Simplify $7 - 2 (\frac{2}{3})$ .

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Simplify each term.

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Multiply $- 2 (\frac{2}{3})$ .

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Combine $- 2$ and $\frac{2}{3}$ .

$y = 7 + \frac{- 2 \cdot 2}{3}$

$x = \frac{2}{3}$

Multiply $- 2$ by $2$ .

$y = 7 + \frac{- 4}{3}$

$x = \frac{2}{3}$

$y = 7 + \frac{- 4}{3}$

$x = \frac{2}{3}$

Move the negative in front of the fraction.

$y = 7 - \frac{4}{3}$

$x = \frac{2}{3}$

$y = 7 - \frac{4}{3}$

$x = \frac{2}{3}$

To write $7$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = 7 \cdot \frac{3}{3} - \frac{4}{3}$

$x = \frac{2}{3}$

Combine $7$ and $\frac{3}{3}$ .

$y = \frac{7 \cdot 3}{3} - \frac{4}{3}$

$x = \frac{2}{3}$

Combine the numerators over the common denominator.

$y = \frac{7 \cdot 3 - 4}{3}$

$x = \frac{2}{3}$

Simplify the numerator.

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Multiply $7$ by $3$ .

$y = \frac{21 - 4}{3}$

$x = \frac{2}{3}$

Subtract $4$ from $21$ .

$y = \frac{17}{3}$

$x = \frac{2}{3}$

$y = \frac{17}{3}$

$x = \frac{2}{3}$

$y = \frac{17}{3}$

$x = \frac{2}{3}$

$y = \frac{17}{3}$

$x = \frac{2}{3}$

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

$(\frac{2}{3}, \frac{17}{3})$

The result can be shown in multiple forms.

Point Form:

$(\frac{2}{3}, \frac{17}{3})$

Equation Form:

$x = \frac{2}{3}, y = \frac{17}{3}$

Solve by Substitution 2x+y=7 , 5x+y=9

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