Solve {l}{5{(22*5-x)}+9*3x=350}{y=10}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { 5 {(22 \cdot 5 - x)} + 9 \cdot 3 x = 350 }\\ { y = 10 }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

Solve for x, y, z, a, b

$$b=10$$

Solution Steps

Consider the first equation. Multiply 22 and 5 to get 110.

$$5\left(110-x\right)+9\times 3x=350$$

Use the distributive property to multiply 5 by 110-x.

$$550-5x+9\times 3x=350$$

Multiply 9 and 3 to get 27.

$$550-5x+27x=350$$

Combine -5x and 27x to get 22x.

$$550+22x=350$$

Subtract 550 from both sides.

$$22x=350-550$$

Subtract 550 from 350 to get -200.

$$22x=-200$$

Divide both sides by 22.

$$x=\frac{-200}{22}$$

Reduce the fraction \frac{-200}{22} to lowest terms by extracting and canceling out 2.

$$x=-\frac{100}{11}$$

The system is now solved.

$$x=-\frac{100}{11} y=10 z=10 a=10 b=10$$