Finite Math Examples

$x + y = 22$ , $y = x^{2} - 4 x + 4$

Step 1

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

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Step 1.1

Replace all occurrences of $y$ in $x + y = 22$ with $x^{2} - 4 x + 4$ .

$x + x^{2} - 4 x + 4 = 22$

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

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $x + x^{2} - 4 x + 4$ .

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Step 1.2.1.1

Remove parentheses.

$x + x^{2} - 4 x + 4 = 22$

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

Step 1.2.1.2

Subtract $4 x$ from $x$ .

$x^{2} - 3 x + 4 = 22$

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

$x^{2} - 3 x + 4 = 22$

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

$x^{2} - 3 x + 4 = 22$

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

$x^{2} - 3 x + 4 = 22$

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

Step 2

Solve for $x$ in $x^{2} - 3 x + 4 = 22$ .

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Step 2.1

Subtract $22$ from both sides of the equation.

$x^{2} - 3 x + 4 - 22 = 0$

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

Step 2.2

Subtract $22$ from $4$ .

$x^{2} - 3 x - 18 = 0$

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

Step 2.3

Factor $x^{2} - 3 x - 18$ using the AC method.

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Step 2.3.1

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 $- 18$ and whose sum is $- 3$ .

$- 6, 3$

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

Step 2.3.2

Write the factored form using these integers.

$(x - 6) (x + 3) = 0$

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

$(x - 6) (x + 3) = 0$

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

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 $0$ .

$x - 6 = 0$

$x + 3 = 0$

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

Step 2.5

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

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Step 2.5.1

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

$x - 6 = 0$

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

Step 2.5.2

Add $6$ to both sides of the equation.

$x = 6$

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

$x = 6$

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

Step 2.6

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

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

Step 2.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = - 3$

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

Step 2.7

The final solution is all the values that make $(x - 6) (x + 3) = 0$ true.

$x = 6, - 3$

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

$x = 6, - 3$

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

Step 3

Replace all occurrences of $x$ with $6$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 4$ with $6$ .

$y = {(6)}^{2} - 4 \cdot 6 + 4$

$x = 6$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify ${(6)}^{2} - 4 \cdot 6 + 4$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Raise $6$ to the power of $2$ .

$y = 36 - 4 \cdot 6 + 4$

$x = 6$

Step 3.2.1.1.2

Multiply $- 4$ by $6$ .

$y = 36 - 24 + 4$

$x = 6$

$y = 36 - 24 + 4$

$x = 6$

Step 3.2.1.2

Simplify by adding and subtracting.

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Step 3.2.1.2.1

Subtract $24$ from $36$ .

$y = 12 + 4$

$x = 6$

Step 3.2.1.2.2

Add $12$ and $4$ .

$y = 16$

$x = 6$

$y = 16$

$x = 6$

$y = 16$

$x = 6$

$y = 16$

$x = 6$

$y = 16$

$x = 6$

Step 4

Replace all occurrences of $x$ with $- 3$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 4$ with $- 3$ .

$y = {(- 3)}^{2} - 4 \cdot - 3 + 4$

$x = - 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify ${(- 3)}^{2} - 4 \cdot - 3 + 4$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Raise $- 3$ to the power of $2$ .

$y = 9 - 4 \cdot - 3 + 4$

$x = - 3$

Step 4.2.1.1.2

Multiply $- 4$ by $- 3$ .

$y = 9 + 12 + 4$

$x = - 3$

$y = 9 + 12 + 4$

$x = - 3$

Step 4.2.1.2

Simplify by adding numbers.

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Step 4.2.1.2.1

Add $9$ and $12$ .

$y = 21 + 4$

$x = - 3$

Step 4.2.1.2.2

Add $21$ and $4$ .

$y = 25$

$x = - 3$

$y = 25$

$x = - 3$

$y = 25$

$x = - 3$

$y = 25$

$x = - 3$

$y = 25$

$x = - 3$

Step 5

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

$(6, 16)$

$(- 3, 25)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(6, 16), (- 3, 25)$

Equation Form:

$x = 6, y = 16$

$x = - 3, y = 25$

Step 7