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Question

# Find the following product and verify the result for x = − 1, y = − 2: (3x − 5y) (x + y)

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Solution

## To multiply, we will use distributive law as follows: $\left(3x-5y\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}=3x\left(x+y\right)-5y\left(x+y\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}+3xy-5xy-5{y}^{2}\phantom{\rule{0ex}{0ex}}=3{x}^{2}-2xy-5{y}^{2}$ $\therefore$ $\left(3x-5y\right)\left(x+y\right)=3{x}^{2}-2xy-5{y}^{2}$. Now, we put x = $-$1 and y = $-$2 on both sides to verify the result. $\text{LHS}=\left(3x-5y\right)\left(x+y\right)\phantom{\rule{0ex}{0ex}}=\left\{3\left(-1\right)-5\left(-2\right)\right\}\left\{-1+\left(-2\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(-3+10\right)\left(-3\right)\phantom{\rule{0ex}{0ex}}=\left(7\right)\left(-3\right)\phantom{\rule{0ex}{0ex}}=-21$ $\text{RHS}=3{x}^{2}-2xy-5{y}^{2}\phantom{\rule{0ex}{0ex}}=3{\left(-1\right)}^{2}-2\left(-1\right)\left(-2\right)-5{\left(-2\right)}^{2}\phantom{\rule{0ex}{0ex}}=3×1-4-5×4\phantom{\rule{0ex}{0ex}}=3-4-20\phantom{\rule{0ex}{0ex}}=-21\phantom{\rule{0ex}{0ex}}$ Because LHS is equal to RHS, the result is verified. Thus, the answer is $3{x}^{2}-2xy-5{y}^{2}$.

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