Algebra Examples

$y = 2 x^{2} + 6 x - 1$ $y = 2 x + 5$

Step 1

Eliminate the equal sides of each equation and combine.

$2 x^{2} + 6 x - 1 = 2 x + 5$

Step 2

Solve $2 x^{2} + 6 x - 1 = 2 x + 5$ for $x$ .

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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 $2 x$ from both sides of the equation.

$2 x^{2} + 6 x - 1 - 2 x = 5$

Step 2.1.2

Subtract $2 x$ from $6 x$ .

$2 x^{2} + 4 x - 1 = 5$

Step 2.2

Subtract $5$ from both sides of the equation.

$2 x^{2} + 4 x - 1 - 5 = 0$

Step 2.3

Subtract $5$ from $- 1$ .

$2 x^{2} + 4 x - 6 = 0$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $2$ out of $2 x^{2} + 4 x - 6$ .

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Step 2.4.1.1

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

$2 (x^{2}) + 4 x - 6 = 0$

Step 2.4.1.2

Factor $2$ out of $4 x$ .

$2 (x^{2}) + 2 (2 x) - 6 = 0$

Step 2.4.1.3

Factor $2$ out of $- 6$ .

$2 x^{2} + 2 (2 x) + 2 \cdot - 3 = 0$

Step 2.4.1.4

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

$2 (x^{2} + 2 x) + 2 \cdot - 3 = 0$

Step 2.4.1.5

Factor $2$ out of $2 (x^{2} + 2 x) + 2 \cdot - 3$ .

$2 (x^{2} + 2 x - 3) = 0$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} + 2 x - 3$ using the AC method.

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Step 2.4.2.1.1

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 $- 3$ and whose sum is $2$ .

$- 1, 3 + y = 2 x + 5$

Step 2.4.2.1.2

Write the factored form using these integers.

$2 ((x - 1) (x + 3)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$2 (x - 1) (x + 3) = 0$

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 $0$ .

$x - 1 = 0$

$x + 3 = 0 + y = 2 x + 5$

Step 2.6

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

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Step 2.6.1

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

$x - 1 = 0$

Step 2.6.2

Add $1$ to both sides of the equation.

$x = 1$

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 $0$ .

$x + 3 = 0$

Step 2.7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.8

The final solution is all the values that make $2 (x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 3

Evaluate $y$ when $x = 1$ .

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Step 3.1

Substitute $1$ for $x$ .

$y = 2 (1) + 5$

Step 3.2

Simplify $2 (1) + 5$ .

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Step 3.2.1

Multiply $2$ by $1$ .

$y = 2 + 5$

Step 3.2.2

Add $2$ and $5$ .

$y = 7$

Step 4

Evaluate $y$ when $x = - 3$ .

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Step 4.1

Substitute $- 3$ for $x$ .

$y = 2 (- 3) + 5$

Step 4.2

Simplify $2 (- 3) + 5$ .

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Step 4.2.1

Multiply $2$ by $- 3$ .

$y = - 6 + 5$

Step 4.2.2

Add $- 6$ and $5$ .

$y = - 1$

Step 5

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

$(1, 7)$

$(- 3, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, 7), (- 3, - 1)$

Equation Form:

$x = 1, y = 7$

$x = - 3, y = - 1$

Step 7