Linear Algebra Examples

$9 q^{2} - 3 q = 8 q^{2} - 5 q + 63$

Step 1

Move all terms containing $q$ to the left side of the equation.

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Step 1.1

Subtract $8 q^{2}$ from both sides of the equation.

$9 q^{2} - 3 q - 8 q^{2} = - 5 q + 63$

Step 1.2

Add $5 q$ to both sides of the equation.

$9 q^{2} - 3 q - 8 q^{2} + 5 q = 63$

Step 1.3

Subtract $8 q^{2}$ from $9 q^{2}$ .

$q^{2} - 3 q + 5 q = 63$

Step 1.4

Add $- 3 q$ and $5 q$ .

$q^{2} + 2 q = 63$

Step 2

Subtract $63$ from both sides of the equation.

$q^{2} + 2 q - 63 = 0$

Step 3

Factor $q^{2} + 2 q - 63$ using the AC method.

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Step 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 $- 63$ and whose sum is $2$ .

$- 7, 9$

Step 3.2

Write the factored form using these integers.

$(q - 7) (q + 9) = 0$

Step 4

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

$q - 7 = 0$

$q + 9 = 0$

Step 5

Set $q - 7$ equal to $0$ and solve for $q$ .

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Step 5.1

Set $q - 7$ equal to $0$ .

$q - 7 = 0$

Step 5.2

Add $7$ to both sides of the equation.

$q = 7$

Step 6

Set $q + 9$ equal to $0$ and solve for $q$ .

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Step 6.1

Set $q + 9$ equal to $0$ .

$q + 9 = 0$

Step 6.2

Subtract $9$ from both sides of the equation.

$q = - 9$

Step 7

The final solution is all the values that make $(q - 7) (q + 9) = 0$ true.

$q = 7, - 9$

Step 8

The domain is the set of all valid $q$ values.

${q | q = 7, - 9}$

Step 9