How do you solve the system of equations $4 x^{2} + 9 y^{2} = 72$ $4x ^ { 2} + 9y ^ { 2} = 72$ and $x^{2} - y^{2} = 5$ $x ^ { 2} - y ^ { 2} = 5$ ?

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How do you solve the system of equations $4 x^{2} + 9 y^{2} = 72$ $4x ^ { 2} + 9y ^ { 2} = 72$ and $x^{2} - y^{2} = 5$ $x ^ { 2} - y ^ { 2} = 5$ ?

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Answer 1 · 2017-07-21T02:24:39

See below
$x = \pm 3$ $x=+-3$
$y = \pm 2$ $y=+-2$

Explanation:

To solve the system of equations $4 x^{2} + 9 y^{2} = 72 and x^{2} - y^{2} = 5$ $4x^2+9y^2=72 and x^2−y^2=5$ , the easiest method is substitution.

I would solve the second equation for $x^{2}$ $x^2$
$x^{2} - y^{2} = 5 \Rightarrow x^{2} = y^{2} + 5$ $x^2−y^2=5" "rArr" "x^2=y^2+5$

Substitute this into the first equation
$4 (y^{2} + 5) + 9 y^{2} = 72$ $4(y^2+5)+9y^2=72$
$4 y^{2} + 20 + 9 y^{2} = 72$ $4y^2+20+9y^2=72$
combine to get: $13 y^{2} + 20 = 72$ $13y^2+20=72$
subtract $20$ $20$ from both sides: $13 y^{2} + 20 - 20 = 72 - 20$ $13y^2+20-20=72-20$
$13 y^{2} = 52$ $13y^2=52$
divide both sides by 13: $\frac{13 y^{2}}{13} = \frac{52}{13}$ $(13y^2)/13=52/13$
$y^{2} = 4$ $y^2=4$
$y = \pm 2$ $y=+-2$

Substitute into the other equation
$4 x^{2} + 9 (4) = 72$ $4x^2+9(4)=72$
$4 x^{2} = 36 x^{2} = 9$ $4x^2=36" "x^2=9$
$x = \pm 3$ $x=+-3$

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How do you solve the system of equations $4 x^{2} + 9 y^{2} = 72$ $4x ^ { 2} + 9y ^ { 2} = 72$ and $x^{2} - y^{2} = 5$ $x ^ { 2} - y ^ { 2} = 5$ ?

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Explanation:

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How do you solve the system of equations 4x ^ { 2} + 9y ^ { 2} = 724x2+9y2=724x ^ { 2} + 9y ^ { 2} = 72 and x ^ { 2} - y ^ { 2} = 5x2−y2=5x ^ { 2} - y ^ { 2} = 5?

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Explanation:

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How do you solve the system of equations $4 x^{2} + 9 y^{2} = 72$ $4x ^ { 2} + 9y ^ { 2} = 72$ and $x^{2} - y^{2} = 5$ $x ^ { 2} - y ^ { 2} = 5$ ?