Question

Compare the equation x/3+3/2y+4=2y-3 and lx +my-n=0 and write the value of l,m and n.

Answer 1

l=2
m= –3
n= –42
hope it helps....

Answer 2

The correct answer is l = $\frac{1}{3}$, m = $\frac{-1}{2}$ and n = 7.

Given: The equation = $\frac{x}{3}+\frac{3}{2}y+4=2y-3$.

To Find: The value of l, m and n.

Solution:

$\frac{x}{3}+\frac{3}{2}y+4=2y-3$

$\frac{x}{3}+\frac{3}{2}y-2y+4+3=0$

$\frac{x}{3}+\frac{3-4}{2}y+7=0$

$\frac{x}{3}+\frac{-1}{2}y+7=0$

Now compare the equations.

Comparing the coefficient of x.

$l=\frac{1}{3}$

Comparing the coefficient of y.

$m=\frac{-1}{2}$

Comparing the constants.

$-n=7$

n = -7

Hence, the values of l, m and n are $\frac{1}{3},\frac{-1}{2}$ and -7 respectively.

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