Q46. - Algebra 1

Question

CHALENGE Find the value of $\mathit{k}$ for which each equation is an identity.

a. $\mathbf{k}\left(\mathbf{3}\mathbf{x}-\mathbf{2}\right)=\mathbf{4}-\mathbf{6}\mathbf{x}$                       b. $\mathbf{15}\mathbf{y}-\mathbf{10}+\mathbf{k}=\mathbf{2}\left(\mathbf{ky}-\mathbf{1}\right)-\mathbf{y}$

Step-by-Step Solution

Answer

1. The value of $k$ for which the given equation is an identity is $-2$.
2. The value of $k$ for which the given equation is an identity is 8.

Part a. Step 1. Write the given equation.

The given equation is “$k\left(3x-2\right)=4-6x$”.

Part a. Step 2. Solve the given equation.

Solving,

 

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

 

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

 

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

 

So, $3kx=-6x$ and $-2k=4$

 

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

 

$k=-2$  (Simplify)

 

Similarly,

 

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

 

$k=-2$  (Simplify)

 

So, $k=-2$ is the required value.

Part a. Step 3. Check the obtained value.

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

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

 

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.

Part b. Step 1. Write the given equation.

The given equation is “$15y-10+k=2\left(ky-1\right)-y$”.

Part b. Step 2. Solve the given equation.

Solving,

 

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

 

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

 

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

 

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

 

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

 

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

 

$2k-1=15$  (Simplify)

 

$2k-1+1=15+1$  (Add 1 to both sides)

 

$2k=16$  (Simplify)

 

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

 

$k=8$  (Simplify)

 

Similarly,

 

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

 

$k=8$  (Simplify)

 

So, $k=8$ is the required value.

Part b. Step 3. Check the obtained value.

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

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

 

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.