Precalculus Examples

$x + y = 3$ $y = x^{2} - 4 x + 5$

Step 1

Replace all occurrences of $y$ with $x^{2} - 4 x + 5$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $x + y = 3$ with $x^{2} - 4 x + 5$ .

$x + x^{2} - 4 x + 5 = 3$

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

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $x + x^{2} - 4 x + 5$ .

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Step 1.2.1.1

Remove parentheses.

$x + x^{2} - 4 x + 5 = 3$

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

Step 1.2.1.2

Subtract $4 x$ from $x$ .

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

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

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

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

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

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

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

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

Step 2

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

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Step 2.1

Subtract $3$ from both sides of the equation.

$x^{2} - 3 x + 5 - 3 = 0$

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

Step 2.2

Subtract $3$ from $5$ .

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

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

Step 2.3

Factor $x^{2} - 3 x + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

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

Step 2.3.2

Write the factored form using these integers.

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

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

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

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

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 $0$ .

$x - 2 = 0$

$x - 1 = 0$

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

Step 2.5

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

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Step 2.5.1

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

$x - 2 = 0$

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

Step 2.5.2

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 2$

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

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 $0$ .

$x - 1 = 0$

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

Step 2.6.2

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

Step 2.7

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

$x = 2, 1$

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

$x = 2, 1$

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

Step 3

Replace all occurrences of $x$ with $2$ in each equation.

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Step 3.1

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 5$ with $2$ .

$y = {(2)}^{2} - 4 \cdot 2 + 5$

$x = 2$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify ${(2)}^{2} - 4 \cdot 2 + 5$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

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

$y = 4 - 4 \cdot 2 + 5$

$x = 2$

Step 3.2.1.1.2

Multiply $- 4$ by $2$ .

$y = 4 - 8 + 5$

$x = 2$

$y = 4 - 8 + 5$

$x = 2$

Step 3.2.1.2

Simplify by adding and subtracting.

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Step 3.2.1.2.1

Subtract $8$ from $4$ .

$y = - 4 + 5$

$x = 2$

Step 3.2.1.2.2

Add $- 4$ and $5$ .

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

Step 4

Replace all occurrences of $x$ with $1$ in each equation.

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Step 4.1

Replace all occurrences of $x$ in $y = x^{2} - 4 x + 5$ with $1$ .

$y = {(1)}^{2} - 4 \cdot 1 + 5$

$x = 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify ${(1)}^{2} - 4 \cdot 1 + 5$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

One to any power is one.

$y = 1 - 4 \cdot 1 + 5$

$x = 1$

Step 4.2.1.1.2

Multiply $- 4$ by $1$ .

$y = 1 - 4 + 5$

$x = 1$

$y = 1 - 4 + 5$

$x = 1$

Step 4.2.1.2

Simplify by adding and subtracting.

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Step 4.2.1.2.1

Subtract $4$ from $1$ .

$y = - 3 + 5$

$x = 1$

Step 4.2.1.2.2

Add $- 3$ and $5$ .

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

Step 5

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

$(2, 1)$

$(1, 2)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 1), (1, 2)$

Equation Form:

$x = 2, y = 1$

$x = 1, y = 2$

Step 7