Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 832: 55


The solution of the given expression is $41$.

Work Step by Step

We have to multiply both sides by $20$ to get the simplified form: $\begin{align} & 20\cdot \left( \frac{x+3}{4}-\frac{x+1}{10} \right)=20\cdot \left( \frac{x-2}{5}-1 \right) \\ & 20\cdot \frac{x+3}{4}-20\cdot \frac{x+1}{10}=20\cdot \frac{x-2}{5}-20\cdot 1 \\ & 5\left( x+3 \right)-2\left( x+1 \right)=4\left( x-2 \right)-20 \\ & 5x+15-2x-2=4x-8-20 \end{align}$ And combine the terms on both sides to get the equation: $3x+13=4x-28$ And subtract $4x+13$ from both sides to get the value of x, $\begin{align} & 3x+13-\left( 4x+13 \right)=4x-28-\left( 4x+13 \right) \\ & 3x-4x=-28-13 \\ & -x=-41 \end{align}$ And divide both sides by -1 to get: $\begin{align} & \frac{-x}{-1}=\frac{-41}{-1} \\ & x=41 \end{align}$ Then, to check whether the solution is correct or not, put $x=41$ in the equation: $\begin{align} & \frac{41+3}{4}-\frac{41+1}{10}=\frac{41-2}{5}-1 \\ & \frac{44}{4}-\frac{42}{10}=\frac{39}{5}-1 \end{align}$ And multiply both sides by 20 to get: $\begin{align} & 20\left( \frac{44}{4}-\frac{42}{10} \right)=20\left( \frac{39}{5}-1 \right) \\ & 5\cdot 44-2\cdot 42=4\cdot 39-20 \\ & 220-84=156-20 \\ & 136=136 \end{align}$ Here, the left-hand side is equal to the right-hand side of the equation. When we have the left-hand side equal to the right-hand side of the equation, then the equation is satisfied. Hence, the solution of the given expression is $41$.
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