Calculus Examples

$x = 4 t - 3$ , $y = 3 t + 1$

Step 1

Set up the parametric equation for $x (t)$ to solve the equation for $t$ .

$x = 4 t - 3$

Step 2

Rewrite the equation as $4 t - 3 = x$ .

$4 t - 3 = x$

Step 3

Add $3$ to both sides of the equation.

$4 t = x + 3$

Step 4

Divide each term in $4 t = x + 3$ by $4$ and simplify.

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Step 4.1

Divide each term in $4 t = x + 3$ by $4$ .

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $4$ .

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Step 4.2.1.1

Cancel the common factor.

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

Step 4.2.1.2

Divide $t$ by $1$ .

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

Step 5

Replace $t$ in the equation for $y$ to get the equation in terms of $x$ .

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

Step 6

Simplify $3 (\frac{x}{4} + \frac{3}{4}) + 1$ .

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Step 6.1

Simplify each term.

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Step 6.1.1

Apply the distributive property.

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

Step 6.1.2

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

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

Step 6.1.3

Multiply $3 (\frac{3}{4})$ .

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Step 6.1.3.1

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

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

Step 6.1.3.2

Multiply $3$ by $3$ .

$y = \frac{3 x}{4} + \frac{9}{4} + 1$

Step 6.2

Simplify the expression.

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Step 6.2.1

Write $1$ as a fraction with a common denominator.

$y = \frac{3 x}{4} + \frac{9}{4} + \frac{4}{4}$

Step 6.2.2

Combine the numerators over the common denominator.

$y = \frac{3 x}{4} + \frac{9 + 4}{4}$

Step 6.2.3

Add $9$ and $4$ .

$y = \frac{3 x}{4} + \frac{13}{4}$