Calculus Examples

$y = \frac{4}{3} x$ , $y = 4$ , $y = \frac{1}{16} x^{2}$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

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

Step 1.2

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

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Step 1.2.1

Multiply both sides of the equation by $\frac{3}{4}$ .

$\frac{3}{4} \cdot (\frac{4}{3} x) = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2

Simplify both sides of the equation.

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Step 1.2.2.1

Simplify the left side.

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Step 1.2.2.1.1

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

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Step 1.2.2.1.1.1

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

$\frac{3}{4} \cdot \frac{4 x}{3} = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.1.1.2

Cancel the common factor of $3$ .

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Step 1.2.2.1.1.2.1

Cancel the common factor.

$\frac{3}{4} \cdot \frac{4 x}{3} = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.1.1.2.2

Rewrite the expression.

$\frac{1}{4} \cdot (4 x) = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

$\frac{1}{4} \cdot (4 x) = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.1.1.3

Cancel the common factor of $4$ .

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Step 1.2.2.1.1.3.1

Factor $4$ out of $4 x$ .

$\frac{1}{4} \cdot (4 (x)) = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.1.1.3.2

Cancel the common factor.

$\frac{1}{4} \cdot (4 x) = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.1.1.3.3

Rewrite the expression.

$x = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.2

Simplify the right side.

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Step 1.2.2.2.1

Cancel the common factor of $4$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$x = \frac{3}{4} \cdot 4 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.2.2.2.1.2

Rewrite the expression.

$x = 3 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = 3 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = 3 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = 3 + y = 4$

$y = \frac{1}{16} x^{2}$

$x = 3 + y = 4$

$y = \frac{1}{16} x^{2}$

Step 1.3

Substitute $3$ for $x$ .

$y = 4 y = \frac{1}{16} \cdot {(3)}^{2}$

Step 1.4

Substitute $3$ for $x$ in $y = \frac{1}{16} \cdot {(3)}^{2}$ and solve for $y$ .

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Step 1.4.1

Remove parentheses.

$y = \frac{1}{16} \cdot 3^{2}$

Step 1.4.2

Simplify $\frac{1}{16} \cdot 3^{2}$ .

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Step 1.4.2.1

Raise $3$ to the power of $2$ .

$y = \frac{1}{16} \cdot 9$

Step 1.4.2.2

Combine $\frac{1}{16}$ and $9$ .

$y = \frac{9}{16}$

Step 1.5

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

$(3, \frac{9}{16})$

Step 2

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

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