Algebra Examples

$y = 2 x^{2}$ $y = - 2 x + 4$

Step 1

Eliminate the equal sides of each equation and combine.

$2 x^{2} = - 2 x + 4$

Step 2

Solve $2 x^{2} = - 2 x + 4$ for $x$ .

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Step 2.1

Add $2 x$ to both sides of the equation.

$2 x^{2} + 2 x = 4$

Step 2.2

Subtract $4$ from both sides of the equation.

$2 x^{2} + 2 x - 4 = 0$

Step 2.3

Factor the left side of the equation.

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Step 2.3.1

Factor $2$ out of $2 x^{2} + 2 x - 4$ .

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Step 2.3.1.1

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) + 2 x - 4 = 0$

Step 2.3.1.2

Factor $2$ out of $2 x$ .

$2 (x^{2}) + 2 (x) - 4 = 0$

Step 2.3.1.3

Factor $2$ out of $- 4$ .

$2 (x^{2}) + 2 x + 2 \cdot - 2 = 0$

Step 2.3.1.4

Factor $2$ out of $2 (x^{2}) + 2 x$ .

$2 (x^{2} + x) + 2 \cdot - 2 = 0$

Step 2.3.1.5

Factor $2$ out of $2 (x^{2} + x) + 2 \cdot - 2$ .

$2 (x^{2} + x - 2) = 0$

Step 2.3.2

Factor.

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Step 2.3.2.1

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

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Step 2.3.2.1.1

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 $- 2$ and whose sum is $1$ .

$- 1, 2 + y = - 2 x + 4$

Step 2.3.2.1.2

Write the factored form using these integers.

$2 ((x - 1) (x + 2)) = 0$

Step 2.3.2.2

Remove unnecessary parentheses.

$2 (x - 1) (x + 2) = 0$

Step 2.4

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

$x - 1 = 0$

$x + 2 = 0 + y = - 2 x + 4$

Step 2.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 2.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.6

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

The final solution is all the values that make $2 (x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 3

Evaluate $y$ when $x = 1$ .

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Step 3.1

Substitute $1$ for $x$ .

$y = - 2 \cdot 1 + 4$

Step 3.2

Substitute $1$ for $x$ in $y = - 2 \cdot 1 + 4$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = - 2 \cdot 1 + 4$

Step 3.2.2

Simplify $- 2 \cdot 1 + 4$ .

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Step 3.2.2.1

Multiply $- 2$ by $1$ .

$y = - 2 + 4$

Step 3.2.2.2

Add $- 2$ and $4$ .

$y = 2$

Step 4

Evaluate $y$ when $x = - 2$ .

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Step 4.1

Substitute $- 2$ for $x$ .

$y = - 2 \cdot - 2 + 4$

Step 4.2

Substitute $- 2$ for $x$ in $y = - 2 \cdot - 2 + 4$ and solve for $y$ .

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Step 4.2.1

Remove parentheses.

$y = - 2 \cdot - 2 + 4$

Step 4.2.2

Simplify $- 2 \cdot - 2 + 4$ .

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Step 4.2.2.1

Multiply $- 2$ by $- 2$ .

$y = 4 + 4$

Step 4.2.2.2

Add $4$ and $4$ .

$y = 8$

Step 5

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

$(1, 2)$

$(- 2, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, 2), (- 2, 8)$

Equation Form:

$x = 1, y = 2$

$x = - 2, y = 8$

Step 7