F+2=H-10k(E+2)-এর সমাধান করুন

$$F + 2 = H - 10 k ( E + 2 )$$

Solve for E

$$\left\{\begin{matrix}E=\frac{-F+H-20k-2}{10k}\text{, }&k\neq 0\\E\in \mathrm{R}\text{, }&F=H-2\text{ and }k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$H-10k\left(E+2\right)=F+2$$

Use the distributive property to multiply -10k by E+2.

$$H-10kE-20k=F+2$$

Subtract H from both sides.

$$-10kE-20k=F+2-H$$

Add 20k to both sides.

$$-10kE=F+2-H+20k$$

The equation is in standard form.

$$\left(-10k\right)E=F-H+20k+2$$

Divide both sides by -10k.

$$\frac{\left(-10k\right)E}{-10k}=\frac{F-H+20k+2}{-10k}$$

Dividing by -10k undoes the multiplication by -10k.

$$E=\frac{F-H+20k+2}{-10k}$$

Divide F-H+2+20k by -10k.

$$E=-\frac{F-H+20k+2}{10k}$$

Solve for F

$$F=-10Ek+H-20k-2$$

Solution Steps

Subtract 2 from both sides.

$$F=H-10k\left(E+2\right)-2$$

Use the distributive property to multiply -10k by E+2.

$$F=H-10kE-20k-2$$