Solve the System of Equations y=2+1/4x 3x-5y=-3

$y = 2 + \frac{1}{4} x$ $3 x - 5 y = - 3$

Step 1

Simplify $2 + \frac{1}{4} x$ .

Combine $\frac{1}{4}$ and $x$ .

$y = 2 + \frac{x}{4}$

$3 x - 5 y = - 3$

Reorder $2$ and $\frac{x}{4}$ .

$y = \frac{x}{4} + 2$

$3 x - 5 y = - 3$

$y = \frac{x}{4} + 2$

$3 x - 5 y = - 3$

Step 2

Replace all occurrences of $y$ with $\frac{x}{4} + 2$ in each equation.

Replace all occurrences of $y$ in $3 x - 5 y = - 3$ with $\frac{x}{4} + 2$ .

$3 x - 5 (\frac{x}{4} + 2) = - 3$

$y = \frac{x}{4} + 2$

Simplify the left side.

Simplify $3 x - 5 (\frac{x}{4} + 2)$ .

Simplify each term.

Apply the distributive property.

$3 x - 5 \frac{x}{4} - 5 \cdot 2 = - 3$

$y = \frac{x}{4} + 2$

Combine $- 5$ and $\frac{x}{4}$ .

$3 x + \frac{- 5 x}{4} - 5 \cdot 2 = - 3$

$y = \frac{x}{4} + 2$

Multiply $- 5$ by $2$ .

$3 x + \frac{- 5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Move the negative in front of the fraction.

$3 x - \frac{5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$3 x - \frac{5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

To write $3 x$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$3 x \cdot \frac{4}{4} - \frac{5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Simplify terms.

Combine $3 x$ and $\frac{4}{4}$ .

$\frac{3 x \cdot 4}{4} - \frac{5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Combine the numerators over the common denominator.

$\frac{3 x \cdot 4 - 5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{3 x \cdot 4 - 5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $3 x \cdot 4 - 5 x$ .

Factor $x$ out of $3 x \cdot 4$ .

$\frac{x (3 \cdot 4) - 5 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Factor $x$ out of $- 5 x$ .

$\frac{x (3 \cdot 4) + x \cdot - 5}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Factor $x$ out of $x (3 \cdot 4) + x \cdot - 5$ .

$\frac{x (3 \cdot 4 - 5)}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{x (3 \cdot 4 - 5)}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Multiply $3$ by $4$ .

$\frac{x (12 - 5)}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Subtract $5$ from $12$ .

$\frac{x \cdot 7}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{x \cdot 7}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Move $7$ to the left of $x$ .

$\frac{7 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{7 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{7 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{7 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

$\frac{7 x}{4} - 10 = - 3$

$y = \frac{x}{4} + 2$

Step 3

Solve for $x$ in $\frac{7 x}{4} - 10 = - 3$ .

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

Add $10$ to both sides of the equation.

$\frac{7 x}{4} = - 3 + 10$

$y = \frac{x}{4} + 2$

Add $- 3$ and $10$ .

$\frac{7 x}{4} = 7$

$y = \frac{x}{4} + 2$

$\frac{7 x}{4} = 7$

$y = \frac{x}{4} + 2$

Multiply both sides of the equation by $\frac{4}{7}$ .

$\frac{4}{7} \cdot \frac{7 x}{4} = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{4}{7} \cdot \frac{7 x}{4}$ .

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4}{7} \cdot \frac{7 x}{4} = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Rewrite the expression.

$\frac{1}{7} \cdot (7 x) = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

$\frac{1}{7} \cdot (7 x) = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$\frac{1}{7} \cdot (7 (x)) = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Cancel the common factor.

$\frac{1}{7} \cdot (7 x) = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Rewrite the expression.

$x = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

$x = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

$x = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

$x = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Simplify the right side.

Cancel the common factor of $7$ .

Cancel the common factor.

$x = \frac{4}{7} \cdot 7$

$y = \frac{x}{4} + 2$

Rewrite the expression.

$x = 4$

$y = \frac{x}{4} + 2$

$x = 4$

$y = \frac{x}{4} + 2$

$x = 4$

$y = \frac{x}{4} + 2$

$x = 4$

$y = \frac{x}{4} + 2$

$x = 4$

$y = \frac{x}{4} + 2$

Step 4

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

Replace all occurrences of $x$ in $y = \frac{x}{4} + 2$ with $4$ .

$y = \frac{4}{4} + 2$

$x = 4$

Simplify the right side.

Simplify $\frac{4}{4} + 2$ .

Divide $4$ by $4$ .

$y = 1 + 2$

$x = 4$

Add $1$ and $2$ .

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

Step 5

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

$(4, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 3)$

Equation Form:

$x = 4, y = 3$

Step 7

Solve the System of Equations y=2+1/4x 3x-5y=-3

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