Solve a=1/3 b=2/3 (2a-b)^2

Question

Solve the system of equations

(a_{1}, b_{1}) = (0, 0) (a_{2}, b_{2}) = (\frac{3}{2}, \frac{9}{2})

Evaluate

{a = \frac{1}{3} b \frac{1}{3} b = \frac{2}{3} (2 a - b)^{2}

Substitute the given value of a into the equation \frac{1}{3} b = \frac{2}{3} (2 a - b)^{2}

\frac{1}{3} b = \frac{2}{3} (2 \times \frac{1}{3} b - b)^{2}

Simplify

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\frac{1}{3} b = \frac{2}{27} b^{2}

Add or subtract both sides

\frac{1}{3} b - \frac{2}{27} b^{2} = 0

Factor the expression

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\frac{1}{27} b (9 - 2 b) = 0

When the product of factors equals 0,at least one factor is 0

\frac{1}{27} b = 0 9 - 2 b = 0

Solve the equation for b

b = 0 9 - 2 b = 0

Solve the equation for b

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b = 0 b = \frac{9}{2}

Calculate

b = 0 \cup b = \frac{9}{2}

Rearrange the terms

{a = \frac{1}{3} b b = 0 \cup {a = \frac{1}{3} b b = \frac{9}{2}

Calculate

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{a = 0 b = 0 \cup {a = \frac{1}{3} b b = \frac{9}{2}

Calculate

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{a = 0 b = 0 \cup {a = \frac{3}{2} b = \frac{9}{2}

Check the solution

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{a = 0 b = 0 \cup {a = \frac{3}{2} b = \frac{9}{2}

Check the solution

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{a = 0 b = 0 \cup {a = \frac{3}{2} b = \frac{9}{2}

Solution

(a_{1}, b_{1}) = (0, 0) (a_{2}, b_{2}) = (\frac{3}{2}, \frac{9}{2})

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