Question

Multiply 2/3x³y³ by (3x-15y) and verify the result for x= 2, y= -1. proper way​

Answer 1

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Answer 2

Question:

Multiply:-

$\sf \frac{ \frac{2}{3} {x}^{3} {y}^{3} }{3x - 15y}$

and verify the result for x = 2, y = -1 .

Answer:

$\therefore The \: value \: is \: \underline{ \:\underline{ \: \sf \pink{\bold{ \frac{ - 16}{63}}} \: } \: }$

Given:

★ An expression is given,

$\frac{ \frac{2}{3} {x}^{3} {y}^{3} }{3x - 15y}$

$\bigstar The \: value \: of \: x = 2,y = - 1$

Step-by-step explanation:

$\frac{ \frac{2}{3} {x}^{3} {y}^{3} }{3x - 15y}$

$\implies \frac{2 \times {x}^{3} \times {y}^{3} }{3(3x - 15y)}$

$\implies \frac{2 {x}^{3} {y}^{3} }{9x - 45y}$

Now substituting the values of x and y

$\implies \frac{2 \times {(2)}^{3} \times {( - 1)}^{3} }{9(2) - 45( - 1)}$

$\implies \frac{2 \times 8 \times - 1}{18 + 45}$

$\implies \boxed{ \sf\frac{ - 16}{63}}$

$\therefore The \: value \: is \: \underline{ \: \bold{ \frac{ - 16}{63}} \: }$

Verification:

$\sf \frac{ \frac{2}{3} {x}^{3} {y}^{3} }{3x - 15y} = \frac{ - 16}{63}$

To prove L.H.S = R.H.S

$\implies \frac{ \frac{2}{3} \times {(2)}^{3} \times {( - 1)}^{3} }{3(2) - 15( - 1)}$

$\implies \frac{ \frac{2 \times 8 \times - 1}{3} }{6 + 15}$

$\implies \frac{ - 16}{3(6 + 15)}$

$\implies \frac{ - 16}{18 + 15}$

$\implies \boxed{ \sf\frac{ - 16}{63}}$

$\therefore \underline{ \: \bold{L.H.S = R.H.S } \: }$

Hence verified.