Algebra Examples

$y = 2 x^{2} + x$ $2 x + y = 20$

Step 1

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

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Step 1.1

Replace all occurrences of $y$ in $2 x + y = 20$ with $2 x^{2} + x$ .

$2 x + 2 x^{2} + x = 20$

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

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $2 x + 2 x^{2} + x$ .

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Step 1.2.1.1

Remove parentheses.

$2 x + 2 x^{2} + x = 20$

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

Step 1.2.1.2

Add $2 x$ and $x$ .

$2 x^{2} + 3 x = 20$

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

$2 x^{2} + 3 x = 20$

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

$2 x^{2} + 3 x = 20$

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

$2 x^{2} + 3 x = 20$

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

Step 2

Solve for $x$ in $2 x^{2} + 3 x = 20$ .

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Step 2.1

Subtract $20$ from both sides of the equation.

$2 x^{2} + 3 x - 20 = 0$

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

Step 2.2

Factor by grouping.

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Step 2.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 20 = - 40$ and whose sum is $b = 3$ .

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Step 2.2.1.1

Factor $3$ out of $3 x$ .

$2 x^{2} + 3 (x) - 20 = 0$

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

Step 2.2.1.2

Rewrite $3$ as $- 5$ plus $8$

$2 x^{2} + (- 5 + 8) x - 20 = 0$

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

Step 2.2.1.3

Apply the distributive property.

$2 x^{2} - 5 x + 8 x - 20 = 0$

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

$2 x^{2} - 5 x + 8 x - 20 = 0$

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

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 5 x) + 8 x - 20 = 0$

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

Step 2.2.2.2

Factor out the greatest common factor (GCF) from each group.

$x (2 x - 5) + 4 (2 x - 5) = 0$

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

$x (2 x - 5) + 4 (2 x - 5) = 0$

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

Step 2.2.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$(2 x - 5) (x + 4) = 0$

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

$(2 x - 5) (x + 4) = 0$

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

Step 2.3

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

$2 x - 5 = 0$

$x + 4 = 0$

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

Step 2.4

Set $2 x - 5$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

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

Step 2.4.2

Solve $2 x - 5 = 0$ for $x$ .

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Step 2.4.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

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

Step 2.4.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

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Step 2.4.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

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

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

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

Step 2.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

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

$x = \frac{5}{2}$

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

$x = \frac{5}{2}$

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

$x = \frac{5}{2}$

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

$x = \frac{5}{2}$

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

$x = \frac{5}{2}$

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

Step 2.5

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

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Step 2.5.1

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

$x + 4 = 0$

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

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = - 4$

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

Step 2.6

The final solution is all the values that make $(2 x - 5) (x + 4) = 0$ true.

$x = \frac{5}{2}, - 4$

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

$x = \frac{5}{2}, - 4$

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

Step 3

Replace all occurrences of $x$ with $\frac{5}{2}$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = 2 x^{2} + x$ with $\frac{5}{2}$ .

$y = 2 {(\frac{5}{2})}^{2} + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2

Simplify $y = 2 {(\frac{5}{2})}^{2} + \frac{5}{2}$ .

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Remove parentheses.

$y = 2 {(\frac{5}{2})}^{2} + \frac{5}{2}$

$x = \frac{5}{2}$

$y = 2 {(\frac{5}{2})}^{2} + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify $2 {(\frac{5}{2})}^{2} + \frac{5}{2}$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Apply the product rule to $\frac{5}{2}$ .

$y = 2 (\frac{5^{2}}{2^{2}}) + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.1.2

Raise $5$ to the power of $2$ .

$y = 2 (\frac{25}{2^{2}}) + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.1.3

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

$y = 2 (\frac{25}{4}) + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.1.4

Cancel the common factor of $2$ .

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Step 3.2.2.1.1.4.1

Factor $2$ out of $4$ .

$y = 2 (\frac{25}{2 (2)}) + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.1.4.2

Cancel the common factor.

$y = 2 (\frac{25}{2 \cdot 2}) + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.1.4.3

Rewrite the expression.

$y = \frac{25}{2} + \frac{5}{2}$

$x = \frac{5}{2}$

$y = \frac{25}{2} + \frac{5}{2}$

$x = \frac{5}{2}$

$y = \frac{25}{2} + \frac{5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.2

Combine fractions.

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Step 3.2.2.1.2.1

Combine the numerators over the common denominator.

$y = \frac{25 + 5}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.2.2

Simplify the expression.

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Step 3.2.2.1.2.2.1

Add $25$ and $5$ .

$y = \frac{30}{2}$

$x = \frac{5}{2}$

Step 3.2.2.1.2.2.2

Divide $30$ by $2$ .

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

$y = 15$

$x = \frac{5}{2}$

Step 4

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

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Step 4.1

Replace all occurrences of $x$ in $y = 2 x^{2} + x$ with $- 4$ .

$y = 2 {(- 4)}^{2} - 4$

$x = - 4$

Step 4.2

Simplify $y = 2 {(- 4)}^{2} - 4$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$y = 2 {(- 4)}^{2} - 4$

$x = - 4$

$y = 2 {(- 4)}^{2} - 4$

$x = - 4$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Simplify $2 {(- 4)}^{2} - 4$ .

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Step 4.2.2.1.1

Simplify each term.

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Step 4.2.2.1.1.1

Raise $- 4$ to the power of $2$ .

$y = 2 \cdot 16 - 4$

$x = - 4$

Step 4.2.2.1.1.2

Multiply $2$ by $16$ .

$y = 32 - 4$

$x = - 4$

$y = 32 - 4$

$x = - 4$

Step 4.2.2.1.2

Subtract $4$ from $32$ .

$y = 28$

$x = - 4$

$y = 28$

$x = - 4$

$y = 28$

$x = - 4$

$y = 28$

$x = - 4$

$y = 28$

$x = - 4$

Step 5

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

$(\frac{5}{2}, 15)$

$(- 4, 28)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{5}{2}, 15), (- 4, 28)$

Equation Form:

$x = \frac{5}{2}, y = 15$

$x = - 4, y = 28$

Step 7