Algebra Examples

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

Step 1

Replace all occurrences of $y$ with $x - 3$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $y^{2} = x - 3$ with $x - 3$ .

${(x - 3)}^{2} = x - 3$

$y = x - 3$

Step 1.2

Simplify the left side.

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Step 1.2.1

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

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Step 1.2.1.1

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

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

$y = x - 3$

Step 1.2.1.2

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

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Step 1.2.1.2.1

Apply the distributive property.

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

$y = x - 3$

Step 1.2.1.2.2

Apply the distributive property.

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

$y = x - 3$

Step 1.2.1.2.3

Apply the distributive property.

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

$y = x - 3$

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

$y = x - 3$

Step 1.2.1.3

Simplify and combine like terms.

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Step 1.2.1.3.1

Simplify each term.

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Step 1.2.1.3.1.1

Multiply $x$ by $x$ .

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

$y = x - 3$

Step 1.2.1.3.1.2

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

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

$y = x - 3$

Step 1.2.1.3.1.3

Multiply $- 3$ by $- 3$ .

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

$y = x - 3$

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

$y = x - 3$

Step 1.2.1.3.2

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

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

Step 2

Solve for $x$ in $x^{2} - 6 x + 9 = x - 3$ .

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Step 2.1

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

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Step 2.1.1

Subtract $x$ from both sides of the equation.

$x^{2} - 6 x + 9 - x = - 3$

$y = x - 3$

Step 2.1.2

Subtract $x$ from $- 6 x$ .

$x^{2} - 7 x + 9 = - 3$

$y = x - 3$

$x^{2} - 7 x + 9 = - 3$

$y = x - 3$

Step 2.2

Add $3$ to both sides of the equation.

$x^{2} - 7 x + 9 + 3 = 0$

$y = x - 3$

Step 2.3

Add $9$ and $3$ .

$x^{2} - 7 x + 12 = 0$

$y = x - 3$

Step 2.4

Factor $x^{2} - 7 x + 12$ using the AC method.

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Step 2.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $- 7$ .

$- 4, - 3$

$y = x - 3$

Step 2.4.2

Write the factored form using these integers.

$(x - 4) (x - 3) = 0$

$y = x - 3$

$(x - 4) (x - 3) = 0$

$y = x - 3$

Step 2.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x - 3 = 0$

$y = x - 3$

Step 2.6

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

$y = x - 3$

Step 2.6.2

Add $4$ to both sides of the equation.

$x = 4$

$y = x - 3$

$x = 4$

$y = x - 3$

Step 2.7

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

$y = x - 3$

Step 2.7.2

Add $3$ to both sides of the equation.

$x = 3$

$y = x - 3$

$x = 3$

$y = x - 3$

Step 2.8

The final solution is all the values that make $(x - 4) (x - 3) = 0$ true.

$x = 4, 3$

$y = x - 3$

$x = 4, 3$

$y = x - 3$

Step 3

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

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Step 3.1

Replace all occurrences of $x$ in $y = x - 3$ with $4$ .

$y = (4) - 3$

$x = 4$

Step 3.2

Simplify the right side.

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Step 3.2.1

Subtract $3$ from $4$ .

$y = 1$

$x = 4$

$y = 1$

$x = 4$

$y = 1$

$x = 4$

Step 4

Replace all occurrences of $x$ with $3$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = x - 3$ with $3$ .

$y = (3) - 3$

$x = 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

Subtract $3$ from $3$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 5

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

$(4, 1)$

$(3, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 1), (3, 0)$

Equation Form:

$x = 4, y = 1$

$x = 3, y = 0$

Step 7