Algebra Examples

$(5, 3)$ , $(6, - 9)$ , $(- 4, 1)$

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.

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

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.

$(3) = a {(5)}^{2} + b (5) + c$

$(- 9) = a {(6)}^{2} + b (6) + c$

$(1) = a {(- 4)}^{2} + b (- 4) + c$

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

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Solve for $a$ in the first equation.

$a = - \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$

$- 9 = 36 a + 6 b + c$

$1 = 16 a - 4 b + c$

Replace all occurrences of $a$ with the solution found by solving the last equation for $a$ . In this case, the value substituted is $- \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$ .

$a = - \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$

$- 9 = 36 (- \frac{b}{5} - \frac{c}{25} + \frac{3}{25}) + 6 b + c$

$1 = 16 a - 4 b + c$

Replace all occurrences of $a$ with the solution found by solving the last equation for $a$ . In this case, the value substituted is $- \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$ .

$a = - \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$

$- 9 = 36 (- \frac{b}{5} - \frac{c}{25} + \frac{3}{25}) + 6 b + c$

$1 = 16 (- \frac{b}{5} - \frac{c}{25} + \frac{3}{25}) - 4 b + c$

Solve for $b$ in the second equation.

$a = - \frac{b}{5} - \frac{c}{25} + \frac{3}{25}$

$b = - \frac{11 c}{30} + \frac{111}{10}$

$1 = - \frac{36 b}{5} + \frac{9 c}{25} + \frac{48}{25}$

Replace all occurrences of $b$ with the solution found by solving the last equation for $b$ . In this case, the value substituted is $- \frac{11 c}{30} + \frac{111}{10}$ .

$a = - \frac{- \frac{11 c}{30} + \frac{111}{10}}{5} - \frac{c}{25} + \frac{3}{25}$

$b = - \frac{11 c}{30} + \frac{111}{10}$

$1 = - \frac{36 b}{5} + \frac{9 c}{25} + \frac{48}{25}$

Replace all occurrences of $b$ with the solution found by solving the last equation for $b$ . In this case, the value substituted is $- \frac{11 c}{30} + \frac{111}{10}$ .

$a = - \frac{- \frac{11 c}{30} + \frac{111}{10}}{5} - \frac{c}{25} + \frac{3}{25}$

$b = - \frac{11 c}{30} + \frac{111}{10}$

$1 = - \frac{36 (- \frac{11 c}{30} + \frac{111}{10})}{5} + \frac{9 c}{25} + \frac{48}{25}$

Solve for $c$ in the third equation.

$a = \frac{c}{30} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 c}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $\frac{79}{3}$ .

$a = \frac{\frac{79}{3}}{30} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 c}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Replace all occurrences of $c$ with the solution found by solving the last equation for $c$ . In this case, the value substituted is $\frac{79}{3}$ .

$a = \frac{\frac{79}{3}}{30} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify.

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Simplify the right side.

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Simplify each term.

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Multiply the numerator by the reciprocal of the denominator.

$a = \frac{79}{3} \cdot \frac{1}{30} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify $\frac{79}{3} \frac{1}{30}$ .

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Multiply $\frac{79}{3}$ and $\frac{1}{30}$ .

$a = \frac{79}{3 \cdot 30} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $3$ by $30$ .

$a = \frac{79}{90} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{79}{90} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{79}{90} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

To write $\frac{79}{90}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$a = \frac{79}{90} \cdot \frac{5}{5} - \frac{111}{50} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

To write $\frac{111}{50}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$a = \frac{79}{90} \cdot \frac{5}{5} - \frac{111}{50} \cdot \frac{9}{9} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Write each expression with a common denominator of $450$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$a = \frac{79 \cdot 5}{90 \cdot 5} - \frac{111}{50} \cdot \frac{9}{9} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $90$ by $5$ .

$a = \frac{79 \cdot 5}{450} - \frac{111}{50} \cdot \frac{9}{9} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Combine.

$a = \frac{79 \cdot 5}{450} - \frac{111 \cdot 9}{50 \cdot 9} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $50$ by $9$ .

$a = \frac{79 \cdot 5}{450} - \frac{111 \cdot 9}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{79 \cdot 5}{450} - \frac{111 \cdot 9}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Combine the numerators over the common denominator.

$a = \frac{79 \cdot 5 - (111 \cdot 9)}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify the numerator.

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Multiply $79$ by $5$ .

$a = \frac{395 - (111 \cdot 9)}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $111$ by $9$ .

$a = \frac{395 - 1 \cdot 999}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $- 1$ by $999$ .

$a = \frac{395 - 999}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Subtract $999$ from $395$ .

$a = \frac{- 604}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{- 604}{450} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify each term.

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Reduce the expression by cancelling the common factors.

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Factor $2$ out of $450$ .

$a = \frac{2 \cdot - 302}{2 \cdot 225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Cancel the common factor.

$a = \frac{2 \cdot - 302}{2 \cdot 225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Rewrite the expression.

$a = \frac{- 302}{225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{- 302}{225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Move the negative in front of the fraction.

$a = - \frac{302}{225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{302}{225} + \frac{3}{25}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

To write $\frac{3}{25}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$a = - \frac{302}{225} + \frac{3}{25} \cdot \frac{9}{9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Write each expression with a common denominator of $225$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$a = - \frac{302}{225} + \frac{3 \cdot 9}{25 \cdot 9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $25$ by $9$ .

$a = - \frac{302}{225} + \frac{3 \cdot 9}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{302}{225} + \frac{3 \cdot 9}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Combine the numerators over the common denominator.

$a = \frac{- 1 \cdot 302 + 3 \cdot 9}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify the numerator.

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Multiply $- 1$ by $302$ .

$a = \frac{- 302 + 3 \cdot 9}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $3$ by $9$ .

$a = \frac{- 302 + 27}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Add $- 302$ and $27$ .

$a = \frac{- 275}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{- 275}{225}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Reduce the expression by cancelling the common factors.

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Factor $25$ out of $225$ .

$a = \frac{25 \cdot - 11}{25 \cdot 9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Cancel the common factor.

$a = \frac{25 \cdot - 11}{25 \cdot 9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Rewrite the expression.

$a = \frac{- 11}{9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = \frac{- 11}{9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Move the negative in front of the fraction.

$a = - \frac{11}{9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - \frac{11 (\frac{79}{3})}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify the right side.

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Simplify each term.

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Simplify the numerator.

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Write $11$ as a fraction with denominator $1$ .

$a = - \frac{11}{9}$

$b = - \frac{\frac{11}{1} \cdot \frac{79}{3}}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $\frac{11}{1}$ and $\frac{79}{3}$ .

$a = - \frac{11}{9}$

$b = - \frac{\frac{11 \cdot 79}{3}}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - \frac{\frac{11 \cdot 79}{3}}{30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply the numerator by the reciprocal of the denominator.

$a = - \frac{11}{9}$

$b = - (\frac{11 \cdot 79}{3} \cdot \frac{1}{30}) + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $11$ by $79$ .

$a = - \frac{11}{9}$

$b = - (\frac{869}{3} \cdot \frac{1}{30}) + \frac{111}{10}$

$c = \frac{79}{3}$

Simplify $\frac{869}{3} \frac{1}{30}$ .

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Multiply $\frac{869}{3}$ and $\frac{1}{30}$ .

$a = - \frac{11}{9}$

$b = - \frac{869}{3 \cdot 30} + \frac{111}{10}$

$c = \frac{79}{3}$

Multiply $3$ by $30$ .

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111}{10}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111}{10}$

$c = \frac{79}{3}$

To write $\frac{111}{10}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111}{10} \cdot \frac{9}{9}$

$c = \frac{79}{3}$

Write each expression with a common denominator of $90$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111 \cdot 9}{10 \cdot 9}$

$c = \frac{79}{3}$

Multiply $10$ by $9$ .

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111 \cdot 9}{90}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = - \frac{869}{90} + \frac{111 \cdot 9}{90}$

$c = \frac{79}{3}$

Combine the numerators over the common denominator.

$a = - \frac{11}{9}$

$b = \frac{- 1 \cdot 869 + 111 \cdot 9}{90}$

$c = \frac{79}{3}$

Simplify the numerator.

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Multiply $- 1$ by $869$ .

$a = - \frac{11}{9}$

$b = \frac{- 869 + 111 \cdot 9}{90}$

$c = \frac{79}{3}$

Multiply $111$ by $9$ .

$a = - \frac{11}{9}$

$b = \frac{- 869 + 999}{90}$

$c = \frac{79}{3}$

Add $- 869$ and $999$ .

$a = - \frac{11}{9}$

$b = \frac{130}{90}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{130}{90}$

$c = \frac{79}{3}$

Reduce the expression by cancelling the common factors.

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Factor $10$ out of $90$ .

$a = - \frac{11}{9}$

$b = \frac{10 \cdot 13}{10 \cdot 9}$

$c = \frac{79}{3}$

Cancel the common factor.

$a = - \frac{11}{9}$

$b = \frac{10 \cdot 13}{10 \cdot 9}$

$c = \frac{79}{3}$

Rewrite the expression.

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

List the solutions to the system of equations.

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

$a = - \frac{11}{9}$

$b = \frac{13}{9}$

$c = \frac{79}{3}$

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = - \frac{11 x^{2}}{9} + \frac{13 x}{9} + \frac{79}{3}$

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