Solve a b=7 a-b=3 - ScanMath

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Question

Solve the system of equations

(a_{1}, b_{1}) = (\frac{3 + 93}{14}, \frac{- 3 + 93}{2}) (a_{2}, b_{2}) = (\frac{3 - 93}{14}, - \frac{3 + 93}{2})

Evaluate

{ab = 7 a - b 7 a - b = 3

Solve the equation for b

More Steps

{ab = 7 a - b b = - 3 + 7 a

Substitute the given value of b into the equation ab = 7 a - b

a (- 3 + 7 a) = 7 a - (- 3 + 7 a)

Simplify

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a (- 3 + 7 a) = 3

Expand the expression

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- 3 a + 7 a^{2} = 3

Move the expression to the left side

- 3 a + 7 a^{2} - 3 = 0

Rewrite in standard form

7 a^{2} - 3 a - 3 = 0

Substitute a= 7,b= - 3 and c= - 3 into the quadratic formula a = \frac{- b \pm b ^{2} - 4 a c}{2 a}

a = \frac{3 \pm ( - 3 ) ^{2} - 4 \times 7 ( - 3 )}{2 \times 7}

Simplify the expression

a = \frac{3 \pm ( - 3 ) ^{2} - 4 \times 7 ( - 3 )}{14}

Simplify the expression

More Steps

a = \frac{3 \pm 93}{14}

Separate the equation into 2 possible cases

a = \frac{3 + 93}{14} a = \frac{3 - 93}{14}

Evaluate the logic

a = \frac{3 + 93}{14} \cup a = \frac{3 - 93}{14}

Rearrange the terms

{a = \frac{3 + 93}{14} b = - 3 + 7 a \cup {a = \frac{3 - 93}{14} b = - 3 + 7 a

Calculate

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{a = \frac{3 + 93}{14} b = \frac{- 3 + 93}{2} \cup {a = \frac{3 - 93}{14} b = - 3 + 7 a

Calculate

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{a = \frac{3 + 93}{14} b = \frac{- 3 + 93}{2} \cup {a = \frac{3 - 93}{14} b = - \frac{3 + 93}{2}

Check the solution

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{a = \frac{3 + 93}{14} b = \frac{- 3 + 93}{2} \cup {a = \frac{3 - 93}{14} b = - \frac{3 + 93}{2}

Check the solution

More Steps

{a = \frac{3 + 93}{14} b = \frac{- 3 + 93}{2} \cup {a = \frac{3 - 93}{14} b = - \frac{3 + 93}{2}

Solution

(a_{1}, b_{1}) = (\frac{3 + 93}{14}, \frac{- 3 + 93}{2}) (a_{2}, b_{2}) = (\frac{3 - 93}{14}, - \frac{3 + 93}{2})

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