Algebra Examples

$y = 2 x^{2} - 6 x + 3$ , $y = 5 x - 2$

Step 1

Eliminate the equal sides of each equation and combine.

$2 x^{2} - 6 x + 3 = 5 x - 2$

Step 2

Solve $2 x^{2} - 6 x + 3 = 5 x - 2$ 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 $5 x$ from both sides of the equation.

$2 x^{2} - 6 x + 3 - 5 x = - 2$

Step 2.1.2

Subtract $5 x$ from $- 6 x$ .

$2 x^{2} - 11 x + 3 = - 2$

Step 2.2

Add $2$ to both sides of the equation.

$2 x^{2} - 11 x + 3 + 2 = 0$

Step 2.3

Add $3$ and $2$ .

$2 x^{2} - 11 x + 5 = 0$

Step 2.4

Factor by grouping.

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Step 2.4.1

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 $a \cdot c = 2 \cdot 5 = 10$ and whose sum is $b = - 11$ .

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Step 2.4.1.1

Factor $- 11$ out of $- 11 x$ .

$2 x^{2} - 11 x + 5 = 0$

Step 2.4.1.2

Rewrite $- 11$ as $- 1$ plus $- 10$

$2 x^{2} + (- 1 - 10) x + 5 = 0$

Step 2.4.1.3

Apply the distributive property.

$2 x^{2} - 1 x - 10 x + 5 = 0$

Step 2.4.2

Factor out the greatest common factor from each group.

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Step 2.4.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) - 10 x + 5 = 0$

Step 2.4.2.2

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

$x (2 x - 1) - 5 (2 x - 1) = 0$

Step 2.4.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x - 5) = 0$

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 $0$ .

$2 x - 1 = 0$

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

Step 2.6

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 2.6.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 2.6.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 2.6.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 2.6.2.2.1

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.6.2.2.2

Simplify the left side.

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Step 2.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 2.7

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

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Step 2.7.1

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

$x - 5 = 0$

Step 2.7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.8

The final solution is all the values that make $(2 x - 1) (x - 5) = 0$ true.

$x = \frac{1}{2}, 5$

Step 3

Evaluate $y$ when $x = \frac{1}{2}$ .

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Step 3.1

Substitute $\frac{1}{2}$ for $x$ .

$y = 5 (\frac{1}{2}) - 2$

Step 3.2

Simplify $5 (\frac{1}{2}) - 2$ .

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Step 3.2.1

Combine $5$ and $\frac{1}{2}$ .

$y = \frac{5}{2} - 2$

Step 3.2.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = \frac{5}{2} - 2 \cdot \frac{2}{2}$

Step 3.2.3

Combine $- 2$ and $\frac{2}{2}$ .

$y = \frac{5}{2} + \frac{- 2 \cdot 2}{2}$

Step 3.2.4

Combine the numerators over the common denominator.

$y = \frac{5 - 2 \cdot 2}{2}$

Step 3.2.5

Simplify the numerator.

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Step 3.2.5.1

Multiply $- 2$ by $2$ .

$y = \frac{5 - 4}{2}$

Step 3.2.5.2

Subtract $4$ from $5$ .

$y = \frac{1}{2}$

Step 4

Evaluate $y$ when $x = 5$ .

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Step 4.1

Substitute $5$ for $x$ .

$y = 5 (5) - 2$

Step 4.2

Simplify $5 (5) - 2$ .

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Step 4.2.1

Multiply $5$ by $5$ .

$y = 25 - 2$

Step 4.2.2

Subtract $2$ from $25$ .

$y = 23$

Step 5

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

$(\frac{1}{2}, \frac{1}{2})$

$(5, 23)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{2}, \frac{1}{2}), (5, 23)$

Equation Form:

$x = \frac{1}{2}, y = \frac{1}{2}$

$x = 5, y = 23$

Step 7