Finite Math Examples

$p = - q^{2} + 8400$ , $p = 50 q$

Step 1

Eliminate the equal sides of each equation and combine.

$- q^{2} + 8400 = 50 q$

Step 2

Solve $- q^{2} + 8400 = 50 q$ for $q$ .

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Step 2.1

Subtract $50 q$ from both sides of the equation.

$- q^{2} + 8400 - 50 q = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $- q^{2} + 8400 - 50 q$ .

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Step 2.2.1.1

Move $8400$ .

$- q^{2} - 50 q + 8400 = 0$

Step 2.2.1.2

Factor $- 1$ out of $- q^{2}$ .

$- (q^{2}) - 50 q + 8400 = 0$

Step 2.2.1.3

Factor $- 1$ out of $- 50 q$ .

$- (q^{2}) - (50 q) + 8400 = 0$

Step 2.2.1.4

Rewrite $8400$ as $- 1 (- 8400)$ .

$- (q^{2}) - (50 q) - 1 \cdot - 8400 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (q^{2}) - (50 q)$ .

$- (q^{2} + 50 q) - 1 \cdot - 8400 = 0$

Step 2.2.1.6

Factor $- 1$ out of $- (q^{2} + 50 q) - 1 (- 8400)$ .

$- (q^{2} + 50 q - 8400) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $q^{2} + 50 q - 8400$ using the AC method.

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Step 2.2.2.1.1

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 $- 8400$ and whose sum is $50$ .

$- 70, 120 + p = 50 q$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((q - 70) (q + 120)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (q - 70) (q + 120) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$q - 70 = 0$

$q + 120 = 0 + p = 50 q$

Step 2.4

Set $q - 70$ equal to $0$ and solve for $q$ .

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Step 2.4.1

Set $q - 70$ equal to $0$ .

$q - 70 = 0$

Step 2.4.2

Add $70$ to both sides of the equation.

$q = 70$

Step 2.5

Set $q + 120$ equal to $0$ and solve for $q$ .

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Step 2.5.1

Set $q + 120$ equal to $0$ .

$q + 120 = 0$

Step 2.5.2

Subtract $120$ from both sides of the equation.

$q = - 120$

Step 2.6

The final solution is all the values that make $- (q - 70) (q + 120) = 0$ true.

$q = 70, - 120$

Step 3

Evaluate $p$ when $q = 70$ .

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Step 3.1

Substitute $70$ for $q$ .

$p = 50 (70)$

Step 3.2

Multiply $50$ by $70$ .

$p = 3500$

Step 4

Evaluate $p$ when $q = - 120$ .

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Step 4.1

Substitute $- 120$ for $q$ .

$p = 50 (- 120)$

Step 4.2

Multiply $50$ by $- 120$ .

$p = - 6000$

Step 5

The solution of the system of equations is all the values that make the system true.

$p = 3500, q = 70 p = - 6000, q = - 120$

Step 6

List all of the solutions.

$p = 3500, q = 70$

$p = - 6000, q = - 120$