Algebra Examples

$(1, 0)$ ,

$(4, 2)$ ,

$(3, - 7)$

Step 1

Use the standard form of a quadratic equation

$y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Step 2

Create a system of equations by substituting the

$x$ and

$y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$0 = a {(1)}^{2} + b (1) + c, 2 = a {(4)}^{2} + b (4) + c, - 7 = a {(3)}^{2} + b (3) + c$

Step 3

Solve the system of equations to find the values of

$a$ ,

$b$ , and

$c$ .

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Step 3.1

Solve for

$a$ in

$0 = a + b + c$ .

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Step 3.1.1

Rewrite the equation as

$a + b + c = 0$ .

$a + b + c = 0$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2

Move all terms not containing

$a$ to the right side of the equation.

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Step 3.1.2.1

Subtract

$b$ from both sides of the equation.

$a + c = - b$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.1.2.2

Subtract

$c$ from both sides of the equation.

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$a = - b - c$

$2 = a \cdot 4^{2} + b (4) + c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2

Replace all occurrences of

$a$ with

$- b - c$ in each equation.

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Step 3.2.1

Replace all occurrences of

$a$ in

$2 = a \cdot 4^{2} + b (4) + c$ with

$- b - c$ .

$2 = (- b - c) \cdot 4^{2} + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Simplify

$(- b - c) \cdot 4^{2} + b (4) + c$ .

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Step 3.2.2.1.1

Simplify each term.

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Step 3.2.2.1.1.1

Raise

$4$ to the power of

$2$ .

$2 = (- b - c) \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.2

Apply the distributive property.

$2 = - b \cdot 16 - c \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.3

Multiply

$16$ by

$- 1$ .

$2 = - 16 b - c \cdot 16 + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.4

Multiply

$16$ by

$- 1$ .

$2 = - 16 b - 16 c + b (4) + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.1.5

Move

$4$ to the left of

$b$ .

$2 = - 16 b - 16 c + 4 b + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 16 b - 16 c + 4 b + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.2

Simplify by adding terms.

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Step 3.2.2.1.2.1

Add

$- 16 b$ and

$4 b$ .

$2 = - 12 b - 16 c + c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.2.1.2.2

Add

$- 16 c$ and

$c$ .

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = a \cdot 3^{2} + b (3) + c$

Step 3.2.3

Replace all occurrences of

$a$ in

$- 7 = a \cdot 3^{2} + b (3) + c$ with

$- b - c$ .

$- 7 = (- b - c) \cdot 3^{2} + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

Simplify

$(- b - c) \cdot 3^{2} + b (3) + c$ .

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Step 3.2.4.1.1

Simplify each term.

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Step 3.2.4.1.1.1

Raise

$3$ to the power of

$2$ .

$- 7 = (- b - c) \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.2

Apply the distributive property.

$- 7 = - b \cdot 9 - c \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.3

Multiply

$9$ by

$- 1$ .

$- 7 = - 9 b - c \cdot 9 + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.4

Multiply

$9$ by

$- 1$ .

$- 7 = - 9 b - 9 c + b (3) + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.1.5

Move

$3$ to the left of

$b$ .

$- 7 = - 9 b - 9 c + 3 b + c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 9 b - 9 c + 3 b + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.2

Simplify by adding terms.

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Step 3.2.4.1.2.1

Add

$- 9 b$ and

$3 b$ .

$- 7 = - 6 b - 9 c + c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.2.4.1.2.2

Add

$- 9 c$ and

$c$ .

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

$- 7 = - 6 b - 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3

Solve for

$b$ in

$- 7 = - 6 b - 8 c$ .

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Step 3.3.1

Rewrite the equation as

$- 6 b - 8 c = - 7$ .

$- 6 b - 8 c = - 7$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.2

Add

$8 c$ to both sides of the equation.

$- 6 b = - 7 + 8 c$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3

Divide each term in

$- 6 b = - 7 + 8 c$ by

$- 6$ and simplify.

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Step 3.3.3.1

Divide each term in

$- 6 b = - 7 + 8 c$ by

$- 6$ .

$\frac{- 6 b}{- 6} = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of

$- 6$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.2.1.2

Divide

$b$ by

$1$ .

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{- 7}{- 6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Simplify each term.

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Step 3.3.3.3.1.1

Dividing two negative values results in a positive value.

$b = \frac{7}{6} + \frac{8 c}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2

Cancel the common factor of

$8$ and

$- 6$ .

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Step 3.3.3.3.1.2.1

Factor

$2$ out of

$8 c$ .

$b = \frac{7}{6} + \frac{2 (4 c)}{- 6}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2

Cancel the common factors.

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Step 3.3.3.3.1.2.2.1

Factor

$2$ out of

$- 6$ .

$b = \frac{7}{6} + \frac{2 (4 c)}{2 (- 3)}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2.2

Cancel the common factor.

$b = \frac{7}{6} + \frac{2 (4 c)}{2 \cdot - 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.2.2.3

Rewrite the expression.

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} + \frac{4 c}{- 3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.3.3.3.1.3

Move the negative in front of the fraction.

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$2 = - 12 b - 15 c$

$a = - b - c$

Step 3.4

Replace all occurrences of

$b$ with

$\frac{7}{6} - \frac{4 c}{3}$ in each equation.

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Step 3.4.1

Replace all occurrences of

$b$ in

$2 = - 12 b - 15 c$ with

$\frac{7}{6} - \frac{4 c}{3}$ .

$2 = - 12 (\frac{7}{6} - \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify

$- 12 (\frac{7}{6} - \frac{4 c}{3}) - 15 c$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$2 = - 12 (\frac{7}{6}) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2

Cancel the common factor of

$6$ .

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Step 3.4.2.1.1.2.1

Factor

$6$ out of

$- 12$ .

$2 = 6 (- 2) (\frac{7}{6}) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2.2

Cancel the common factor.

$2 = 6 \cdot (- 2 (\frac{7}{6})) - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.2.3

Rewrite the expression.

$2 = - 2 \cdot 7 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 2 \cdot 7 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.3

Multiply

$- 2$ by

$7$ .

$2 = - 14 - 12 (- \frac{4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4

Cancel the common factor of

$3$ .

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Step 3.4.2.1.1.4.1

Move the leading negative in

$- \frac{4 c}{3}$ into the numerator.

$2 = - 14 - 12 \frac{- 4 c}{3} - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.2

Factor

$3$ out of

$- 12$ .

$2 = - 14 + 3 (- 4) (\frac{- 4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.3

Cancel the common factor.

$2 = - 14 + 3 \cdot (- 4 \frac{- 4 c}{3}) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.4.4

Rewrite the expression.

$2 = - 14 - 4 (- 4 c) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 - 4 (- 4 c) - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.1.5

Multiply

$- 4$ by

$- 4$ .

$2 = - 14 + 16 c - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + 16 c - 15 c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.2.1.2

Subtract

$15 c$ from

$16 c$ .

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - b - c$

Step 3.4.3

Replace all occurrences of

$b$ in

$a = - b - c$ with

$\frac{7}{6} - \frac{4 c}{3}$ .

$a = - (\frac{7}{6} - \frac{4 c}{3}) - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4

Simplify the right side.

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Step 3.4.4.1

Simplify

$- (\frac{7}{6} - \frac{4 c}{3}) - c$ .

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Step 3.4.4.1.1

Simplify each term.

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Step 3.4.4.1.1.1

Apply the distributive property.

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.1.2

Multiply

$- - \frac{4 c}{3}$ .

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Step 3.4.4.1.1.2.1

Multiply

$- 1$ by

$- 1$ .

$a = - \frac{7}{6} + 1 (\frac{4 c}{3}) - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.1.2.2

Multiply

$\frac{4 c}{3}$ by

$1$ .

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c}{3} - c$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.2

To write

$- c$ as a fraction with a common denominator, multiply by

$\frac{3}{3}$ .

$a = - \frac{7}{6} + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.3

Simplify terms.

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Step 3.4.4.1.3.1

Combine

$- c$ and

$\frac{3}{3}$ .

$a = - \frac{7}{6} + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.3.2

Combine the numerators over the common denominator.

$a = - \frac{7}{6} + \frac{4 c - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{4 c - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4

Simplify each term.

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Step 3.4.4.1.4.1

Simplify the numerator.

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Step 3.4.4.1.4.1.1

Factor

$c$ out of

$4 c - c \cdot 3$ .

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Step 3.4.4.1.4.1.1.1

Factor

$c$ out of

$4 c$ .

$a = - \frac{7}{6} + \frac{c \cdot 4 - c \cdot 3}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.1.2

Factor

$c$ out of

$- c \cdot 3$ .

$a = - \frac{7}{6} + \frac{c \cdot 4 + c (- 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.1.3

Factor

$c$ out of

$c \cdot 4 + c (- 1 \cdot 3)$ .

$a = - \frac{7}{6} + \frac{c (4 - 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c (4 - 1 \cdot 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.2

Multiply

$- 1$ by

$3$ .

$a = - \frac{7}{6} + \frac{c (4 - 3)}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.1.3

Subtract

$3$ from

$4$ .

$a = - \frac{7}{6} + \frac{c \cdot 1}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c \cdot 1}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.4.4.1.4.2

Multiply

$c$ by

$1$ .

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{c}{3}$

$2 = - 14 + c$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5

Solve for

$c$ in

$2 = - 14 + c$ .

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Step 3.5.1

Rewrite the equation as

$- 14 + c = 2$ .

$- 14 + c = 2$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5.2

Move all terms not containing

$c$ to the right side of the equation.

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Step 3.5.2.1

Add

$14$ to both sides of the equation.

$c = 2 + 14$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.5.2.2

Add

$2$ and

$14$ .

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

$c = 16$

$a = - \frac{7}{6} + \frac{c}{3}$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6

Replace all occurrences of

$c$ with

$16$ in each equation.

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Step 3.6.1

Replace all occurrences of

$c$ in

$a = - \frac{7}{6} + \frac{c}{3}$ with

$16$ .

$a = - \frac{7}{6} + \frac{16}{3}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify

$- \frac{7}{6} + \frac{16}{3}$ .

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Step 3.6.2.1.1

To write

$\frac{16}{3}$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$a = - \frac{7}{6} + \frac{16}{3} \cdot \frac{2}{2}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.2

Write each expression with a common denominator of

$6$ , by multiplying each by an appropriate factor of

$1$ .

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Step 3.6.2.1.2.1

Multiply

$\frac{16}{3}$ by

$\frac{2}{2}$ .

$a = - \frac{7}{6} + \frac{16 \cdot 2}{3 \cdot 2}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.2.2

Multiply

$3$ by

$2$ .

$a = - \frac{7}{6} + \frac{16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = - \frac{7}{6} + \frac{16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.3

Combine the numerators over the common denominator.

$a = \frac{- 7 + 16 \cdot 2}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.4

Simplify the numerator.

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Step 3.6.2.1.4.1

Multiply

$16$ by

$2$ .

$a = \frac{- 7 + 32}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.2.1.4.2

Add

$- 7$ and

$32$ .

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{4 c}{3}$

Step 3.6.3

Replace all occurrences of

$c$ in

$b = \frac{7}{6} - \frac{4 c}{3}$ with

$16$ .

$b = \frac{7}{6} - \frac{4 (16)}{3}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4

Simplify the right side.

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Step 3.6.4.1

Simplify

$\frac{7}{6} - \frac{4 (16)}{3}$ .

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Step 3.6.4.1.1

Multiply

$4$ by

$16$ .

$b = \frac{7}{6} - \frac{64}{3}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.2

To write

$- \frac{64}{3}$ as a fraction with a common denominator, multiply by

$\frac{2}{2}$ .

$b = \frac{7}{6} - \frac{64}{3} \cdot \frac{2}{2}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.3

Write each expression with a common denominator of

$6$ , by multiplying each by an appropriate factor of

$1$ .

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Step 3.6.4.1.3.1

Multiply

$\frac{64}{3}$ by

$\frac{2}{2}$ .

$b = \frac{7}{6} - \frac{64 \cdot 2}{3 \cdot 2}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.3.2

Multiply

$3$ by

$2$ .

$b = \frac{7}{6} - \frac{64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{7}{6} - \frac{64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.4

Combine the numerators over the common denominator.

$b = \frac{7 - 64 \cdot 2}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.5

Simplify the numerator.

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Step 3.6.4.1.5.1

Multiply

$- 64$ by

$2$ .

$b = \frac{7 - 128}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.5.2

Subtract

$128$ from

$7$ .

$b = \frac{- 121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = \frac{- 121}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.6.4.1.6

Move the negative in front of the fraction.

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

$b = - \frac{121}{6}$

$a = \frac{25}{6}$

$c = 16$

Step 3.7

List all of the solutions.

$b = - \frac{121}{6}, a = \frac{25}{6}, c = 16$

Step 4

Substitute the actual values of

$a$ ,

$b$ , and

$c$ into the formula for a quadratic equation to find the resulting equation.

$y = \frac{25 x^{2}}{6} - \frac{121 x}{6} + 16$

Step 5

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