Solve by Substitution x+3=y , 3x+4y=7

$x + 3 = y$ , $3 x + 4 y = 7$

Step 1

Subtract $3$ from both sides of the equation.

$x = y - 3$

$3 x + 4 y = 7$

Step 2

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

Replace all occurrences of $x$ in $3 x + 4 y = 7$ with $y - 3$ .

$3 (y - 3) + 4 y = 7$

$x = y - 3$

Simplify the left side.

Simplify $3 (y - 3) + 4 y$ .

Simplify each term.

Apply the distributive property.

$3 y + 3 \cdot - 3 + 4 y = 7$

$x = y - 3$

Multiply $3$ by $- 3$ .

$3 y - 9 + 4 y = 7$

$x = y - 3$

$3 y - 9 + 4 y = 7$

$x = y - 3$

Add $3 y$ and $4 y$ .

$7 y - 9 = 7$

$x = y - 3$

$7 y - 9 = 7$

$x = y - 3$

$7 y - 9 = 7$

$x = y - 3$

$7 y - 9 = 7$

$x = y - 3$

Step 3

Solve for $y$ in $7 y - 9 = 7$ .

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

Add $9$ to both sides of the equation.

$7 y = 7 + 9$

$x = y - 3$

Add $7$ and $9$ .

$7 y = 16$

$x = y - 3$

$7 y = 16$

$x = y - 3$

Divide each term in $7 y = 16$ by $7$ and simplify.

Divide each term in $7 y = 16$ by $7$ .

$\frac{7 y}{7} = \frac{16}{7}$

$x = y - 3$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{16}{7}$

$x = y - 3$

Divide $y$ by $1$ .

$y = \frac{16}{7}$

$x = y - 3$

$y = \frac{16}{7}$

$x = y - 3$

$y = \frac{16}{7}$

$x = y - 3$

$y = \frac{16}{7}$

$x = y - 3$

$y = \frac{16}{7}$

$x = y - 3$

Step 4

Replace all occurrences of $y$ with $\frac{16}{7}$ in each equation.

Replace all occurrences of $y$ in $x = y - 3$ with $\frac{16}{7}$ .

$x = (\frac{16}{7}) - 3$

$y = \frac{16}{7}$

Simplify the right side.

Simplify $(\frac{16}{7}) - 3$ .

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

$x = \frac{16}{7} - 3 \cdot \frac{7}{7}$

$y = \frac{16}{7}$

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

$x = \frac{16}{7} + \frac{- 3 \cdot 7}{7}$

$y = \frac{16}{7}$

Combine the numerators over the common denominator.

$x = \frac{16 - 3 \cdot 7}{7}$

$y = \frac{16}{7}$

Simplify the numerator.

Multiply $- 3$ by $7$ .

$x = \frac{16 - 21}{7}$

$y = \frac{16}{7}$

Subtract $21$ from $16$ .

$x = \frac{- 5}{7}$

$y = \frac{16}{7}$

$x = \frac{- 5}{7}$

$y = \frac{16}{7}$

Move the negative in front of the fraction.

$x = - \frac{5}{7}$

$y = \frac{16}{7}$

$x = - \frac{5}{7}$

$y = \frac{16}{7}$

$x = - \frac{5}{7}$

$y = \frac{16}{7}$

$x = - \frac{5}{7}$

$y = \frac{16}{7}$

Step 5

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

$(- \frac{5}{7}, \frac{16}{7})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- \frac{5}{7}, \frac{16}{7})$

Equation Form:

$x = - \frac{5}{7}, y = \frac{16}{7}$

Step 7

Solve by Substitution x+3=y , 3x+4y=7

Related Questions

Support LCMGCF.com by disabling your adblocker.