Algebra Examples

$y = x^{2} + 3 x - 7$ $y - 5 x + 8 = 0$

Step 1

Replace all occurrences of $y$ with $x^{2} + 3 x - 7$ in each equation.

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Step 1.1

Replace all occurrences of $y$ in $y - 5 x + 8 = 0$ with $x^{2} + 3 x - 7$ .

$(x^{2} + 3 x - 7) - 5 x + 8 = 0$

$y = x^{2} + 3 x - 7$

Step 1.2

Simplify the left side.

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Step 1.2.1

Simplify $(x^{2} + 3 x - 7) - 5 x + 8$ .

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Step 1.2.1.1

Subtract $5 x$ from $3 x$ .

$x^{2} - 2 x - 7 + 8 = 0$

$y = x^{2} + 3 x - 7$

Step 1.2.1.2

Add $- 7$ and $8$ .

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

$y = x^{2} + 3 x - 7$

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

$y = x^{2} + 3 x - 7$

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

$y = x^{2} + 3 x - 7$

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

$y = x^{2} + 3 x - 7$

Step 2

Solve for $x$ in $x^{2} - 2 x + 1 = 0$ .

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Step 2.1

Factor using the perfect square rule.

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Step 2.1.1

Rewrite $1$ as $1^{2}$ .

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

$y = x^{2} + 3 x - 7$

Step 2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

$y = x^{2} + 3 x - 7$

Step 2.1.3

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 1 + 1^{2} = 0$

$y = x^{2} + 3 x - 7$

Step 2.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

$y = x^{2} + 3 x - 7$

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

$y = x^{2} + 3 x - 7$

Step 2.2

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

$x - 1 = 0$

$y = x^{2} + 3 x - 7$

Step 2.3

Add $1$ to both sides of the equation.

$x = 1$

$y = x^{2} + 3 x - 7$

$x = 1$

$y = x^{2} + 3 x - 7$

Step 3

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

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Step 3.1

Replace all occurrences of $x$ in $y = x^{2} + 3 x - 7$ with $1$ .

$y = {(1)}^{2} + 3 (1) - 7$

$x = 1$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify ${(1)}^{2} + 3 (1) - 7$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

One to any power is one.

$y = 1 + 3 (1) - 7$

$x = 1$

Step 3.2.1.1.2

Multiply $3$ by $1$ .

$y = 1 + 3 - 7$

$x = 1$

$y = 1 + 3 - 7$

$x = 1$

Step 3.2.1.2

Simplify by adding and subtracting.

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Step 3.2.1.2.1

Add $1$ and $3$ .

$y = 4 - 7$

$x = 1$

Step 3.2.1.2.2

Subtract $7$ from $4$ .

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

$y = - 3$

$x = 1$

Step 4

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

$(1, - 3)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(1, - 3)$

Equation Form:

$x = 1, y = - 3$

Step 6