Solve the System of Equations x+y=19 y=3x-1

$x + y = 19$ $y = 3 x - 1$

Step 1

Replace all occurrences of $y$ with $3 x - 1$ in each equation.

Replace all occurrences of $y$ in $x + y = 19$ with $3 x - 1$ .

$x + 3 x - 1 = 19$

$y = 3 x - 1$

Simplify the left side.

Simplify $x + 3 x - 1$ .

Remove parentheses.

$x + 3 x - 1 = 19$

$y = 3 x - 1$

Add $x$ and $3 x$ .

$4 x - 1 = 19$

$y = 3 x - 1$

$4 x - 1 = 19$

$y = 3 x - 1$

$4 x - 1 = 19$

$y = 3 x - 1$

$4 x - 1 = 19$

$y = 3 x - 1$

Step 2

Solve for $x$ in $4 x - 1 = 19$ .

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

Add $1$ to both sides of the equation.

$4 x = 19 + 1$

$y = 3 x - 1$

Add $19$ and $1$ .

$4 x = 20$

$y = 3 x - 1$

$4 x = 20$

$y = 3 x - 1$

Divide each term in $4 x = 20$ by $4$ and simplify.

Divide each term in $4 x = 20$ by $4$ .

$\frac{4 x}{4} = \frac{20}{4}$

$y = 3 x - 1$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{20}{4}$

$y = 3 x - 1$

Divide $x$ by $1$ .

$x = \frac{20}{4}$

$y = 3 x - 1$

$x = \frac{20}{4}$

$y = 3 x - 1$

$x = \frac{20}{4}$

$y = 3 x - 1$

Simplify the right side.

Divide $20$ by $4$ .

$x = 5$

$y = 3 x - 1$

$x = 5$

$y = 3 x - 1$

$x = 5$

$y = 3 x - 1$

$x = 5$

$y = 3 x - 1$

Step 3

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

Replace all occurrences of $x$ in $y = 3 x - 1$ with $5$ .

$y = 3 (5) - 1$

$x = 5$

Simplify the right side.

Simplify $3 (5) - 1$ .

Multiply $3$ by $5$ .

$y = 15 - 1$

$x = 5$

Subtract $1$ from $15$ .

$y = 14$

$x = 5$

$y = 14$

$x = 5$

$y = 14$

$x = 5$

$y = 14$

$x = 5$

Step 4

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

$(5, 14)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(5, 14)$

Equation Form:

$x = 5, y = 14$

Step 6

Solve the System of Equations x+y=19 y=3x-1

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