Solve (41-2j-5k)=(4j-2j-5k)*(4j-2j

$$( 41 - 2 j - 5 k ) = ( 4 j - 2 j - 5 k ) \cdot ( 4 j - 2 j$$

Solve for k

$$k=-\frac{41-2j-4j^{2}}{5\left(2j-1\right)}$$

$$j\neq \frac{1}{2}$$

Steps for Solving Linear Equation

Combine 4j and -2j to get 2j.

$$41-2j-5k=\left(2j-5k\right)\left(4j-2j\right)$$

Combine 4j and -2j to get 2j.

$$41-2j-5k=\left(2j-5k\right)\times 2j$$

Use the distributive property to multiply 2j-5k by 2.

$$41-2j-5k=\left(4j-10k\right)j$$

Use the distributive property to multiply 4j-10k by j.

$$41-2j-5k=4j^{2}-10kj$$

Add 10kj to both sides.

$$41-2j-5k+10kj=4j^{2}$$

Subtract 41 from both sides.

$$-2j-5k+10kj=4j^{2}-41$$

Add 2j to both sides.

$$-5k+10kj=4j^{2}-41+2j$$

Combine all terms containing k.

$$\left(-5+10j\right)k=4j^{2}-41+2j$$

The equation is in standard form.

$$\left(10j-5\right)k=4j^{2}+2j-41$$

Divide both sides by -5+10j.

$$\frac{\left(10j-5\right)k}{10j-5}=\frac{4j^{2}+2j-41}{10j-5}$$

Dividing by -5+10j undoes the multiplication by -5+10j.

$$k=\frac{4j^{2}+2j-41}{10j-5}$$

Divide 4j^{2}-41+2j by -5+10j.

$$k=\frac{4j^{2}+2j-41}{5\left(2j-1\right)}$$

Solve for j

$$j=\frac{-\sqrt{25k^{2}-30k+165}+5k-1}{4}$$

$$j=\frac{\sqrt{25k^{2}-30k+165}+5k-1}{4}$$