Question

Can you help me with this question.

If x+3y = 21 and xy = 22, find the value of x2 +9y2 and (x-3y)2

Answer 1

Answer:

Step-by-step explanation:

To find the value of x2 +9y2, we can start by substituting the second equation into the first equation to solve for y:

x + 3y = 21

xy = 22

Substituting the second equation into the first equation, we get:

x + 3(22/x) = 21

Solving for x, we get:

x = 21 / (1 + 3(22/x))

Multiplying both sides of the equation by x, we get:

x2 = 21x / (1 + 3(22/x))

Multiplying both sides of the equation by x, we get:

x3 = 21x2 / (1 + 3(22/x))

Substituting the second equation into the equation above, we get:

(22)x = 21x2 / (1 + 3(22/x))

Solving for x2, we get:

x2 = (22x) / (1 + 3(22/x))

Substituting the second equation into the equation above, we get:

x2 = (22(22)) / (1 + 3(22/22))

Solving for x2, we get:

x2 = 484 / 4 = 121

Now that we have found the value of x2, we can substitute it back into the second equation to solve for y2:

x2 + 9y2 = 121

xy = 22

Substituting the second equation into the first equation, we get:

121 + 9y2 = 22y

Solving for y2, we get:

y2 = (22y - 121) / 9

Substituting the second equation into the equation above, we get:

y2 = (22(22) - 121) / 9

Solving for y2, we get:

y2 = 121 / 9 = 13.44

So the value of x2 +9y2 is 121 + 9(13.44) = 121 + 120.96 = 241.96

To find the value of (x-3y)2, we can substitute the values of x and y into the equation:

(x-3y)2 = (x2 - 6xy + 9y2)

Substituting the values of x2 and y2 that we found above, we get:

(x-3y)2 = (121 - 6(22) + 9(13.44))

Solving for (x-3y)2, we get:

(x-3y)2 = (-6)2 = 36

So the value of (x-3y)2 is 36.

Answer 2

Answer:

$x^2+9y^2=309$

$(x-3y)^2=177$

Step-by-step explanation:

Given equations:

$\begin{cases}x+3y=21\\xy=22\end{cases}$

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To find the value of x² + 9y²,  first rewrite 9 as 3²:

$\implies x^2+3^2y^2$

$\implies x^2+(3y)^2$

$\boxed{\begin{minipage}{5cm}\underline{Sum of two squares}\\\\ a^2+b^2=(a+b)^2-2ab \\ \end{minipage}}$

Apply the sum of two squares where:

• a = x
• b = 3y

$\begin{aligned}\implies x^2+(3y)^2&=(x+3y)^2-2x(3y)\\&=(x+3y)^2-6xy\end{aligned}$

Substitute the given values of (x + 3y) and xy:

$\begin{aligned}\implies (x+3y)^2-6xy&=21^2-6(22)\\&=441-132\\&=309\end{aligned}$

Therefore:

$\boxed{x^2+9y^2=309}$

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To find the value of (x - 3y)², expand:

$\begin{aligned}\implies (x-3y)^2&=x^2-6xy+9y^2\\&=x^2+9y^2-6xy \end{aligned}$

Rewrite 9 as 3²:

$\begin{aligned}\implies x^2+9y^2-6xy&=x^2+3^2y^2-6xy\\&= x^2+9y^2-6xy\end{aligned}$

$\boxed{\begin{minipage}{5cm}\underline{Sum of two squares}\\\\ a^2+b^2=(a+b)^2-2ab \\ \end{minipage}}$

Apply the sum of two squares  to x² + 9y² where:

• a = x
• b = 3y

$\begin{aligned}\implies x^2+9y^2-6xy&=(x+3y)^2-2x(3y)-6xy\\&=(x+3y)^2-12xy\end{aligned}$

Substitute the given values of (x + 3y) and xy:

$\begin{aligned}\implies (x+3y)^2-12xy&=21^2-12(22)\\&=441-264\\&=177\end{aligned}$

Therefore:

$\boxed{(x-3y)^2=177}$