Pre-Algebra Examples

$\frac{4 t^{2}}{5} = \frac{3 t}{5} + \frac{27}{10}$

Step 1

Multiply both sides by $5$ .

$\frac{4 t^{2}}{5} \cdot 5 = (\frac{3 t}{5} + \frac{27}{10}) \cdot 5$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Cancel the common factor of $5$ .

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Step 2.1.1.1

Cancel the common factor.

$\frac{4 t^{2}}{5} \cdot 5 = (\frac{3 t}{5} + \frac{27}{10}) \cdot 5$

Step 2.1.1.2

Rewrite the expression.

$4 t^{2} = (\frac{3 t}{5} + \frac{27}{10}) \cdot 5$

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $(\frac{3 t}{5} + \frac{27}{10}) \cdot 5$ .

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Step 2.2.1.1

Apply the distributive property.

$4 t^{2} = \frac{3 t}{5} \cdot 5 + \frac{27}{10} \cdot 5$

Step 2.2.1.2

Cancel the common factor of $5$ .

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Step 2.2.1.2.1

Cancel the common factor.

$4 t^{2} = \frac{3 t}{5} \cdot 5 + \frac{27}{10} \cdot 5$

Step 2.2.1.2.2

Rewrite the expression.

$4 t^{2} = 3 t + \frac{27}{10} \cdot 5$

Step 2.2.1.3

Cancel the common factor of $5$ .

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Step 2.2.1.3.1

Factor $5$ out of $10$ .

$4 t^{2} = 3 t + \frac{27}{5 (2)} \cdot 5$

Step 2.2.1.3.2

Cancel the common factor.

$4 t^{2} = 3 t + \frac{27}{5 \cdot 2} \cdot 5$

Step 2.2.1.3.3

Rewrite the expression.

$4 t^{2} = 3 t + \frac{27}{2}$

Step 3

Solve for $t$ .

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Step 3.1

Subtract $3 t$ from both sides of the equation.

$4 t^{2} - 3 t = \frac{27}{2}$

Step 3.2

Subtract $\frac{27}{2}$ from both sides of the equation.

$4 t^{2} - 3 t - \frac{27}{2} = 0$

Step 3.3

Multiply through by the least common denominator $2$ , then simplify.

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Step 3.3.1

Apply the distributive property.

$2 (4 t^{2}) + 2 (- 3 t) + 2 (- \frac{27}{2}) = 0$

Step 3.3.2

Simplify.

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Step 3.3.2.1

Multiply $4$ by $2$ .

$8 t^{2} + 2 (- 3 t) + 2 (- \frac{27}{2}) = 0$

Step 3.3.2.2

Multiply $- 3$ by $2$ .

$8 t^{2} - 6 t + 2 (- \frac{27}{2}) = 0$

Step 3.3.2.3

Cancel the common factor of $2$ .

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Step 3.3.2.3.1

Move the leading negative in $- \frac{27}{2}$ into the numerator.

$8 t^{2} - 6 t + 2 (\frac{- 27}{2}) = 0$

Step 3.3.2.3.2

Cancel the common factor.

$8 t^{2} - 6 t + 2 (\frac{- 27}{2}) = 0$

Step 3.3.2.3.3

Rewrite the expression.

$8 t^{2} - 6 t - 27 = 0$

Step 3.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.5

Substitute the values $a = 8$ , $b = - 6$ , and $c = - 27$ into the quadratic formula and solve for $t$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (8 \cdot - 27)}}{2 \cdot 8}$

Step 3.6

Simplify.

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Step 3.6.1

Simplify the numerator.

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Step 3.6.1.1

Raise $- 6$ to the power of $2$ .

$t = \frac{6 \pm \sqrt{36 - 4 \cdot 8 \cdot - 27}}{2 \cdot 8}$

Step 3.6.1.2

Multiply $- 4 \cdot 8 \cdot - 27$ .

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Step 3.6.1.2.1

Multiply $- 4$ by $8$ .

$t = \frac{6 \pm \sqrt{36 - 32 \cdot - 27}}{2 \cdot 8}$

Step 3.6.1.2.2

Multiply $- 32$ by $- 27$ .

$t = \frac{6 \pm \sqrt{36 + 864}}{2 \cdot 8}$

Step 3.6.1.3

Add $36$ and $864$ .

$t = \frac{6 \pm \sqrt{900}}{2 \cdot 8}$

Step 3.6.1.4

Rewrite $900$ as $30^{2}$ .

$t = \frac{6 \pm \sqrt{30^{2}}}{2 \cdot 8}$

Step 3.6.1.5

Pull terms out from under the radical, assuming positive real numbers.

$t = \frac{6 \pm 30}{2 \cdot 8}$

Step 3.6.2

Multiply $2$ by $8$ .

$t = \frac{6 \pm 30}{16}$

Step 3.6.3

Simplify $\frac{6 \pm 30}{16}$ .

$t = \frac{3 \pm 15}{8}$

Step 3.7

The final answer is the combination of both solutions.

$t = \frac{9}{4}, - \frac{3}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$t = \frac{9}{4}, - \frac{3}{2}$

Decimal Form:

$t = 2.25, - 1.5$

Mixed Number Form:

$t = 2 \frac{1}{4}, - 1 \frac{1}{2}$