Question

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

Answer 1

Question:

Multiply:-

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

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

To find:

★ To find the value of the given expression.

Answer:

$\underline{ \: { \pink{ \frac{ - 16}{63} }} \: } \: is \: the \: value.$

Given:

★ An expression is given.

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,

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

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

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

Verification:

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

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

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

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

$\implies \frac{ - 16}{18 + 45}$

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

Hence verified.