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Expert-verified Found in: Page 689 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Solve the system, or show that it has no solution. If the system has infinitely many solutions,express them in the ordered-pair form given in Example 6. $\begin{array}{l}26x-10y=-4\\ 0.6x+1.2y=-3\end{array}$

Hence, the solution of the system of equations $\begin{array}{l}26x-10y=-4\\ 0.6x+1.2y=-3\end{array}$ is the ordered pair

See the step by step solution

## Step 1. Given information.

Given a system of equations $\begin{array}{l}26x-10y=-4\\ 0.6x+1.2y=-3\end{array}$

substitution method.

## Step 2. Write down the concept.

Given equation,

$26x-10y=-4$ …….(1)

$0.6x+1.2y=-3$ …….(2)

Solving equation (1) for x,

$x=\frac{1}{26}\left(-4+10y\right)$

## Step 3. Determining the angle

We are substituting equation (3) in equation (2),

$-0.6\left(\frac{-2}{13}+\frac{5}{13}y\right)+1.2y=3$

$\frac{6}{65}-\frac{3}{13}y+\frac{12}{10}y=3$

Subtracting $\frac{6}{65}$ from both sides,

$\begin{array}{l}-\frac{3}{13}y+\frac{12}{10}y=3-\frac{6}{65}\\ \frac{-30y+156y}{130}=3-\frac{6}{65}\\ \frac{126y}{130}=\frac{189}{65}\end{array}$

Multiplying both sides by $\frac{130}{126}$

$y=3$

Now, substituting back value of y in equation (3)

$\begin{array}{l}x=\frac{-2}{13}+\frac{5}{13}\left(3\right)\\ x=\frac{-2}{13}+\frac{15}{13}\\ x=1\end{array}$ ### Want to see more solutions like these? 