Pre-Algebra Examples

$y = \frac{4}{3} x$ , $y = \frac{5}{4} x$

Step 1

Eliminate the equal sides of each equation and combine.

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

Step 2

Solve $\frac{4}{3} x = \frac{5}{4} x$ for $x$ .

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Step 2.1

Simplify $\frac{4}{3} x$ .

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Step 2.1.1

Rewrite.

$0 + 0 + \frac{4}{3} x = \frac{5}{4} x$

Step 2.1.2

Simplify by adding zeros.

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

Move all terms containing $x$ to the left side of the equation.

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Step 2.3.1

Subtract $\frac{5 x}{4}$ from both sides of the equation.

$\frac{4 x}{3} - \frac{5 x}{4} = 0$

Step 2.3.2

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

$\frac{4 x}{3} \cdot \frac{4}{4} - \frac{5 x}{4} = 0$

Step 2.3.3

To write $- \frac{5 x}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{4 x}{3} \cdot \frac{4}{4} - \frac{5 x}{4} \cdot \frac{3}{3} = 0$

Step 2.3.4

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.4.1

Multiply $\frac{4 x}{3}$ by $\frac{4}{4}$ .

$\frac{4 x \cdot 4}{3 \cdot 4} - \frac{5 x}{4} \cdot \frac{3}{3} = 0$

Step 2.3.4.2

Multiply $3$ by $4$ .

$\frac{4 x \cdot 4}{12} - \frac{5 x}{4} \cdot \frac{3}{3} = 0$

Step 2.3.4.3

Multiply $\frac{5 x}{4}$ by $\frac{3}{3}$ .

$\frac{4 x \cdot 4}{12} - \frac{5 x \cdot 3}{4 \cdot 3} = 0$

Step 2.3.4.4

Multiply $4$ by $3$ .

$\frac{4 x \cdot 4}{12} - \frac{5 x \cdot 3}{12} = 0$

Step 2.3.5

Combine the numerators over the common denominator.

$\frac{4 x \cdot 4 - 5 x \cdot 3}{12} = 0$

Step 2.3.6

Simplify the numerator.

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Step 2.3.6.1

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

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Step 2.3.6.1.1

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

$\frac{x (4 \cdot 4) - 5 x \cdot 3}{12} = 0$

Step 2.3.6.1.2

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

$\frac{x (4 \cdot 4) + x (- 5 \cdot 3)}{12} = 0$

Step 2.3.6.1.3

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

$\frac{x (4 \cdot 4 - 5 \cdot 3)}{12} = 0$

Step 2.3.6.2

Multiply $4$ by $4$ .

$\frac{x (16 - 5 \cdot 3)}{12} = 0$

Step 2.3.6.3

Multiply $- 5$ by $3$ .

$\frac{x (16 - 15)}{12} = 0$

Step 2.3.6.4

Subtract $15$ from $16$ .

$\frac{x \cdot 1}{12} = 0$

Step 2.3.7

Multiply $x$ by $1$ .

$\frac{x}{12} = 0$

Step 2.4

Set the numerator equal to zero.

$x = 0$

Step 3

Evaluate $y$ when $x = 0$ .

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Step 3.1

Substitute $0$ for $x$ .

$y = \frac{5}{4} \cdot (0)$

Step 3.2

Substitute $0$ for $x$ in $y = \frac{5}{4} \cdot (0)$ and solve for $y$ .

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Step 3.2.1

Multiply $\frac{5}{4}$ by $0$ .

$y = \frac{5}{4} \cdot 0$

Step 3.2.2

Multiply $\frac{5}{4}$ by $0$ .

$y = 0$

Step 4

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

$(0, 0)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(0, 0)$

Equation Form:

$x = 0, y = 0$

Step 6