Precalculus Examples

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ , $y = {(x - 3)}^{2}$

Step 1

Simplify ${(x - 3)}^{2}$ .

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Step 1.1

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$y = (x - 3) (x - 3)$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2

Expand $(x - 3) (x - 3)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$y = x (x - 3) - 3 (x - 3)$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2.2

Apply the distributive property.

$y = x \cdot x + x \cdot - 3 - 3 (x - 3)$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.2.3

Apply the distributive property.

$y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$y = x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.1.2

Move $- 3$ to the left of $x$ .

$y = x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.1.3

Multiply $- 3$ by $- 3$ .

$y = x^{2} - 3 x - 3 x + 9$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$y = x^{2} - 3 x - 3 x + 9$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 1.3.2

Subtract $3 x$ from $- 3 x$ .

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

Step 2

Replace all occurrences of $y$ with $x^{2} - 6 x + 9$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$ with $x^{2} - 6 x + 9$ .

$\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 9)}^{2}}{25}$ .

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Step 2.2.1.1

Simplify the numerator.

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Step 2.2.1.1.1

Factor using the perfect square rule.

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Step 2.2.1.1.1.1

Rewrite $9$ as $3^{2}$ .

$\frac{x^{2}}{9} + \frac{{(x^{2} - 6 x + 3^{2})}^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 2.2.1.1.1.3

Rewrite the polynomial.

$\frac{x^{2}}{9} + \frac{{(x^{2} - 2 \cdot x \cdot 3 + 3^{2})}^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 3$ .

$\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{{({(x - 3)}^{2})}^{2}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.2

Multiply the exponents in ${({(x - 3)}^{2})}^{2}$ .

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Step 2.2.1.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{2 \cdot 2}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.2.2

Multiply $2$ by $2$ .

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{{(x - 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.3

Use the Binomial Theorem.

$\frac{x^{2}}{9} + \frac{x^{4} + 4 x^{3} \cdot - 3 + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4

Simplify each term.

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Step 2.2.1.1.4.1

Multiply $- 3$ by $4$ .

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} {(- 3)}^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.2

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

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 6 x^{2} \cdot 9 + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.3

Multiply $9$ by $6$ .

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x {(- 3)}^{3} + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.4

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

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} + 4 x \cdot - 27 + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.5

Multiply $- 27$ by $4$ .

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + {(- 3)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.4.6

Raise $- 3$ to the power of $4$ .

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{x^{4} - 12 x^{3} + 54 x^{2} - 108 x + 81}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.5

Make each term match the terms from the binomial theorem formula.

$\frac{x^{2}}{9} + \frac{x^{4} - 4 \cdot (3 x^{3}) + 6 \cdot (3^{2} x^{2}) - 4 \cdot (3^{3} x) + 3^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.1.6

Factor $x^{4} - 4 \cdot 3 x^{3} + 6 \cdot 3^{2} x^{2} - 4 \cdot 3^{3} x + 3^{4}$ using the binomial theorem.

$\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2}}{9} + \frac{{(3 - x)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.2

To write $\frac{x^{2}}{9}$ as a fraction with a common denominator, multiply by $\frac{25}{25}$ .

$\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.3

To write $\frac{{(3 - x)}^{4}}{25}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\frac{x^{2}}{9} \cdot \frac{25}{25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.4

Write each expression with a common denominator of $225$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.2.1.4.1

Multiply $\frac{x^{2}}{9}$ by $\frac{25}{25}$ .

$\frac{x^{2} \cdot 25}{9 \cdot 25} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.2

Multiply $9$ by $25$ .

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4}}{25} \cdot \frac{9}{9} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.3

Multiply $\frac{{(3 - x)}^{4}}{25}$ by $\frac{9}{9}$ .

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{25 \cdot 9} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.4.4

Multiply $25$ by $9$ .

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{x^{2} \cdot 25}{225} + \frac{{(3 - x)}^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.5

Combine the numerators over the common denominator.

$\frac{x^{2} \cdot 25 + {(3 - x)}^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6

Simplify the numerator.

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Step 2.2.1.6.1

Move $25$ to the left of $x^{2}$ .

$\frac{25 \cdot x^{2} + {(3 - x)}^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.2

Use the Binomial Theorem.

$\frac{25 x^{2} + (3^{4} + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3

Simplify each term.

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Step 2.2.1.6.3.1

Raise $3$ to the power of $4$ .

$\frac{25 x^{2} + (81 + 4 \cdot (3^{3} (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.2

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

$\frac{25 x^{2} + (81 + 4 \cdot (27 (- x)) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.3

Multiply $4$ by $27$ .

$\frac{25 x^{2} + (81 + 108 (- x) + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.4

Multiply $- 1$ by $108$ .

$\frac{25 x^{2} + (81 - 108 x + 6 \cdot (3^{2} {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.5

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

$\frac{25 x^{2} + (81 - 108 x + 6 \cdot (9 {(- x)}^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.6

Multiply $6$ by $9$ .

$\frac{25 x^{2} + (81 - 108 x + 54 {(- x)}^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.7

Apply the product rule to $- x$ .

$\frac{25 x^{2} + (81 - 108 x + 54 ({(- 1)}^{2} x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.8

Raise $- 1$ to the power of $2$ .

$\frac{25 x^{2} + (81 - 108 x + 54 (1 x^{2}) + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.9

Multiply $x^{2}$ by $1$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 4 \cdot (3 {(- x)}^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.10

Multiply $4$ by $3$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 {(- x)}^{3} + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.11

Apply the product rule to $- x$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 ({(- 1)}^{3} x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.12

Raise $- 1$ to the power of $3$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} + 12 (- x^{3}) + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.13

Multiply $- 1$ by $12$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- x)}^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.14

Apply the product rule to $- x$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + {(- 1)}^{4} x^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.15

Raise $- 1$ to the power of $4$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + 1 x^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.3.16

Multiply $x^{4}$ by $1$ .

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{25 x^{2} + (81 - 108 x + 54 x^{2} - 12 x^{3} + x^{4}) \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.4

Apply the distributive property.

$\frac{25 x^{2} + 81 \cdot 9 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5

Simplify.

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Step 2.2.1.6.5.1

Multiply $81$ by $9$ .

$\frac{25 x^{2} + 729 - 108 x \cdot 9 + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.2

Multiply $9$ by $- 108$ .

$\frac{25 x^{2} + 729 - 972 x + 54 x^{2} \cdot 9 - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.3

Multiply $9$ by $54$ .

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 12 x^{3} \cdot 9 + x^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.4

Multiply $9$ by $- 12$ .

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + x^{4} \cdot 9}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.5.5

Move $9$ to the left of $x^{4}$ .

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{25 x^{2} + 729 - 972 x + 486 x^{2} - 108 x^{3} + 9 \cdot x^{4}}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.6

Add $25 x^{2}$ and $486 x^{2}$ .

$\frac{511 x^{2} + 729 - 972 x - 108 x^{3} + 9 x^{4}}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 2.2.1.6.7

Reorder terms.

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

$y = x^{2} - 6 x + 9$

$\frac{9 x^{4} - 108 x^{3} + 511 x^{2} - 972 x + 729}{225} = 1$

$y = x^{2} - 6 x + 9$

Step 3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 0.80535522, 3$

$y = x^{2} - 6 x + 9$

Step 4

Remove any equations from the system that are always true.

$x \approx 0.80535522$

$y = x^{2} - 6 x + 9$

Step 5