Algebra Examples

$y = 3 x^{2} - 7 x + 1$ $y = - 4 x + 19$

Step 1

Eliminate the equal sides of each equation and combine.

$3 x^{2} - 7 x + 1 = - 4 x + 19$

Step 2

Solve $3 x^{2} - 7 x + 1 = - 4 x + 19$ 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

Add $4 x$ to both sides of the equation.

$3 x^{2} - 7 x + 1 + 4 x = 19$

Step 2.1.2

Add $- 7 x$ and $4 x$ .

$3 x^{2} - 3 x + 1 = 19$

Step 2.2

Subtract $19$ from both sides of the equation.

$3 x^{2} - 3 x + 1 - 19 = 0$

Step 2.3

Subtract $19$ from $1$ .

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

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor $3$ out of $3 x^{2} - 3 x - 18$ .

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Step 2.4.1.1

Factor $3$ out of $3 x^{2}$ .

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

Step 2.4.1.2

Factor $3$ out of $- 3 x$ .

$3 (x^{2}) + 3 (- x) - 18 = 0$

Step 2.4.1.3

Factor $3$ out of $- 18$ .

$3 (x^{2}) + 3 (- x) + 3 (- 6) = 0$

Step 2.4.1.4

Factor $3$ out of $3 (x^{2}) + 3 (- x)$ .

$3 (x^{2} - x) + 3 (- 6) = 0$

Step 2.4.1.5

Factor $3$ out of $3 (x^{2} - x) + 3 (- 6)$ .

$3 (x^{2} - x - 6) = 0$

Step 2.4.2

Factor.

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Step 2.4.2.1

Factor $x^{2} - x - 6$ using the AC method.

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Step 2.4.2.1.1

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 $- 6$ and whose sum is $- 1$ .

$- 3, 2 + y = - 4 x + 19$

Step 2.4.2.1.2

Write the factored form using these integers.

$3 ((x - 3) (x + 2)) = 0$

Step 2.4.2.2

Remove unnecessary parentheses.

$3 (x - 3) (x + 2) = 0$

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 $0$ .

$x - 3 = 0$

$x + 2 = 0 + y = - 4 x + 19$

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 $0$ .

$x - 3 = 0$

Step 2.6.2

Add $3$ to both sides of the equation.

$x = 3$

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

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.8

The final solution is all the values that make $3 (x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Step 3

Evaluate $y$ when $x = 3$ .

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Step 3.1

Substitute $3$ for $x$ .

$y = - 4 \cdot 3 + 19$

Step 3.2

Substitute $3$ for $x$ in $y = - 4 \cdot 3 + 19$ and solve for $y$ .

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Step 3.2.1

Remove parentheses.

$y = - 4 \cdot 3 + 19$

Step 3.2.2

Simplify $- 4 \cdot 3 + 19$ .

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Step 3.2.2.1

Multiply $- 4$ by $3$ .

$y = - 12 + 19$

Step 3.2.2.2

Add $- 12$ and $19$ .

$y = 7$

Step 4

Evaluate $y$ when $x = - 2$ .

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Step 4.1

Substitute $- 2$ for $x$ .

$y = - 4 \cdot - 2 + 19$

Step 4.2

Substitute $- 2$ for $x$ in $y = - 4 \cdot - 2 + 19$ and solve for $y$ .

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Step 4.2.1

Remove parentheses.

$y = - 4 \cdot - 2 + 19$

Step 4.2.2

Simplify $- 4 \cdot - 2 + 19$ .

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Step 4.2.2.1

Multiply $- 4$ by $- 2$ .

$y = 8 + 19$

Step 4.2.2.2

Add $8$ and $19$ .

$y = 27$

Step 5

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

$(3, 7)$

$(- 2, 27)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 7), (- 2, 27)$

Equation Form:

$x = 3, y = 7$

$x = - 2, y = 27$

Step 7