Solve by Substitution y=x+5 , 4x+y=20

$y = x + 5$ , $4 x + y = 20$

Replace all occurrences of $y$ with $x + 5$ in each equation.

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Replace all occurrences of $y$ in $4 x + y = 20$ with $x + 5$ .

$4 x + x + 5 = 20$

$y = x + 5$

Simplify $4 x + x + 5$ .

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Remove parentheses.

$4 x + x + 5 = 20$

$y = x + 5$

Add $4 x$ and $x$ .

$5 x + 5 = 20$

$y = x + 5$

$5 x + 5 = 20$

$y = x + 5$

$5 x + 5 = 20$

$y = x + 5$

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 $5$ from both sides of the equation.

$5 x = 20 - 5$

$y = x + 5$

Subtract $5$ from $20$ .

$5 x = 15$

$y = x + 5$

$5 x = 15$

$y = x + 5$

Divide each term by $5$ and simplify.

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

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

$y = x + 5$

Cancel the common factor of $5$ .

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Cancel the common factor.

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

$y = x + 5$

Divide $x$ by $1$ .

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

$y = x + 5$

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

$y = x + 5$

Divide $15$ by $5$ .

$x = 3$

$y = x + 5$

$x = 3$

$y = x + 5$

$x = 3$

$y = x + 5$

Replace all occurrences of $x$ with $3$ in each equation.

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Replace all occurrences of $x$ in $y = x + 5$ with $3$ .

$y = (3) + 5$

$x = 3$

Add $3$ and $5$ .

$y = 8$

$x = 3$

$y = 8$

$x = 3$

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

$(3, 8)$

The result can be shown in multiple forms.

Point Form:

$(3, 8)$

Equation Form:

$x = 3, y = 8$

Solve by Substitution y=x+5 , 4x+y=20