Precalculus Examples

$y = x^{2} - 3 x + 2$ , $y = 5 x - 10$

Step 1

Eliminate the equal sides of each equation and combine.

$x^{2} - 3 x + 2 = 5 x - 10$

Step 2

Solve $x^{2} - 3 x + 2 = 5 x - 10$ 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.

$x^{2} - 3 x + 2 - 5 x = - 10$

Step 2.1.2

Subtract $5 x$ from $- 3 x$ .

$x^{2} - 8 x + 2 = - 10$

Step 2.2

Add $10$ to both sides of the equation.

$x^{2} - 8 x + 2 + 10 = 0$

Step 2.3

Add $2$ and $10$ .

$x^{2} - 8 x + 12 = 0$

Step 2.4

Factor $x^{2} - 8 x + 12$ 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 $- 8$ .

$- 6, - 2 + y = 5 x - 10$

Step 2.4.2

Write the factored form using these integers.

$(x - 6) (x - 2) = 0$

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 $0$ .

$x - 6 = 0$

$x - 2 = 0 + y = 5 x - 10$

Step 2.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.7

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

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Step 2.7.1

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

$x - 2 = 0$

Step 2.7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.8

The final solution is all the values that make $(x - 6) (x - 2) = 0$ true.

$x = 6, 2$

Step 3

Evaluate $y$ when $x = 6$ .

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Step 3.1

Substitute $6$ for $x$ .

$y = 5 (6) - 10$

Step 3.2

Simplify $5 (6) - 10$ .

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Step 3.2.1

Multiply $5$ by $6$ .

$y = 30 - 10$

Step 3.2.2

Subtract $10$ from $30$ .

$y = 20$

Step 4

Evaluate $y$ when $x = 2$ .

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Step 4.1

Substitute $2$ for $x$ .

$y = 5 (2) - 10$

Step 4.2

Simplify $5 (2) - 10$ .

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Step 4.2.1

Multiply $5$ by $2$ .

$y = 10 - 10$

Step 4.2.2

Subtract $10$ from $10$ .

$y = 0$

Step 5

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

$(6, 20)$

$(2, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, 20), (2, 0)$

Equation Form:

$x = 6, y = 20$

$x = 2, y = 0$

Step 7