Solve h=w^(5/2) w=h^(2/5)

Question

Solve the system of equations

Solve using the substitution method

Solve using the elimination method

(h_{1}, w_{1}) = (0, 0) (h_{2}, w_{2}) = (1, 1)

Evaluate

{h = w^{\frac{5}{2}} \times w w^{\frac{5}{2}} \times w = h^{\frac{2}{5}}

Calculate

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{h = w^{\frac{7}{2}} w^{\frac{5}{2}} \times w = h^{\frac{2}{5}}

Calculate

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{h = w^{\frac{7}{2}} w^{\frac{7}{2}} = h^{\frac{2}{5}}

Substitute the given value of h into the equation w^{\frac{7}{2}} = h^{\frac{2}{5}}

w^{\frac{7}{2}} = (w^{\frac{7}{2}})^{\frac{2}{5}}

Simplify

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w^{\frac{7}{2}} = w^{\frac{7}{5}}

Evaluate

w^{\frac{7}{2}} = w^{\frac{7}{5}}, w^{\frac{7}{5}} \geq 0

The only way a base raised to an odd power can be greater than or equal to 0 is if the base is greater than or equal to 0

w^{\frac{7}{2}} = w^{\frac{7}{5}}, w \geq 0

Solve the equation

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w = 0 \cup w = 1, w \geq 0

Calculate

w = 0 \cup w = 1

Rearrange the terms

{h = w^{\frac{7}{2}} w = 0 \cup {h = w^{\frac{7}{2}} w = 1

Calculate

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{h = 0 w = 0 \cup {h = w^{\frac{7}{2}} w = 1

Calculate

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{h = 0 w = 0 \cup {h = 1 w = 1

Check the solution

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{h = 0 w = 0 \cup {h = 1 w = 1

Check the solution

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{h = 0 w = 0 \cup {h = 1 w = 1

Solution

(h_{1}, w_{1}) = (0, 0) (h_{2}, w_{2}) = (1, 1)

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