91影视

The given equation is 鈥�$k\left(3x鈭�2\right)=4鈭�6x$鈥�.

Solving,

$k\left(3x鈭�2\right)=4鈭�6x$ (Given equation)

$3kx鈭�2k=4鈭�6x$ (Distributive property)

For this to be an identity, the terms containing variable and the constant terms on both sides must be equal.

So,$3kx=鈭�6x$ and$鈭�2k=4$

$\frac{3kx}{3x}=\frac{鈭�6x}{3x}$ (Divide both sides by $3x$)

$k=鈭�2$ (Simplify)

Similarly,

$\frac{鈭�2k}{鈭�2}=\frac{4}{鈭�2}$ (Divide both sides by $-2$)

$k=鈭�2$ (Simplify)

So,$k=鈭�2$ is the required value.

Put$k=鈭�2$ in the given equation$k\left(3x鈭�2\right)=4鈭�6x$ to get,

$$\begin{gathered} \left(鈭�2\right)\left(3x鈭�2\right)=4鈭�6x \\ 鈭�6x+4=4鈭�6x \\ 4鈭�6x=4鈭�6x \end{gathered}$$

Since the expressions obtained on the left-hand side and right-hand side are equal, the obtained value is indeed the required value for which the given equation is an identity.

The given equation is 鈥�$15y鈭�10+k=2\left(ky鈭�1\right)鈭�y$鈥�.

Solving,

$15y鈭�10+k=2\left(ky鈭�1\right)鈭�y$ (Given equation)

$15y鈭�10+k=2ky鈭�2鈭�y$ (Distributive property)

$15y鈭�10+k=\left(2k鈭�1\right)y鈭�2$ (Simplify)

For this to be an identity, the terms containing variable and the constant terms on both sides must be equal.

So,$\left(2k鈭�1\right)y=15y$ and$鈭�10+k=鈭�2$

$\frac{\left(2k鈭�1\right)y}{y}=\frac{15y}{y}$ (Divide both sides by $y$)

$2k鈭�1=15$ (Simplify)

$2k鈭�1+1=15+1$ (Add 1 to both sides)

$2k=16$ (Simplify)

$\frac{2k}{2}=\frac{16}{2}$ (Divide both sides by 2)

$k=8$ (Simplify)

Similarly,

$鈭�10+k+10=鈭�2+10$ (Add 10 to both sides)

$k=8$ (Simplify)

So,$k=8$ is the required value.

Put$k=8$ in the given equation$15y鈭�10+k=2\left(ky鈭�1\right)鈭�y$ to get,

$$\begin{gathered} 15y鈭�10+8=2\left(8y鈭�1\right)鈭�y \\ 15y鈭�2=16y鈭�2鈭�y \\ 15y鈭�2=15y鈭�2 \end{gathered}$$

Since the expressions obtained on the left-hand side and right-hand side are equal, the obtained value is indeed the required value for which the given equation is an identity.