Question

Solve the linear inequality $8z+3-2z < 51$

Answer 1

We recall the properties of inequalities when we add, subtract or multiply the same number on both sides of the inequality. We first simplify the left side of the equation adding the variable term in $z$. We then subtract 3 both sides and then divided 6 both sides to find the solution.

Complete step-by-step solution:
We know that an inequality is a mathematical relation which makes a non-equal comparison between two numbers or other mathematical expressions. There are fundamentally two types of inequalities. If $a$ and $b$ are two real numbers then $a < b$ means $a$ is less than $b$ and $a > b$ means $a$ is greater than $b$.$$$$
We know that if we add or subtract the same number on both sides then the inequality holds. It means for any number $c$ we have

& a < b\Rightarrow a+c < b+c \\\ & a > b\Rightarrow a+c > b+c \\\ \end{aligned}$$ We know that if we multiply or divide a positive number the inequality holds. It means for $c>0$we have; $$\begin{aligned} & a < b\Rightarrow a\times c < b\times c \\\ & a > b\Rightarrow \dfrac{a}{c} > \dfrac{b}{c} \\\ \end{aligned}$$ We know that when we are asked to solve an inequality it means we have to all values satisfying the inequality. We are given the following inequality solve $$8z+3-2z < 51$$ We see that we have only variable $z$ and inequality is less than type. We simplify the left hand side by adding variable terms in $z$ to have; $$\begin{aligned} & \Rightarrow 8z-2z+3 < 51 \\\ & \Rightarrow 6z+3 < 51 \\\ \end{aligned}$$ We now subtract 3 both sides of the above step to have $$\begin{aligned} & \Rightarrow 6z+3-3 < 51-3 \\\ & \Rightarrow 6z < 48 \\\ \end{aligned}$$ We divide 6 both sides of the above step to have $$\begin{aligned} & \Rightarrow \dfrac{6z}{6} < \dfrac{48}{6} \\\ & \Rightarrow z < 8 \\\ \end{aligned}$$ We see from the above result that the solutions for $z$ are all real numbers which are less than 8. $$$$ **Note:** We can write the solution for $z$ in interval notation as $\left( -\infty ,8 \right)$. We note that when we multiply or divide a negative number the inequality is reversed which means if $a < 0$ we have $a < b \Rightarrow ac > bc $ and $a < b\Rightarrow \dfrac{a}{c} > \dfrac{b}{c}$. Here in this problem 6 was a positive number and hence dividing by 6 inequality holds.