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Q48.

Expert-verifiedFound in: Page 689

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 $>">1,3$

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

substitution method.

Given equation,

$26x-10y=-4$ …….(1)

**$0.6x+1.2y=-3$ **…….(2)** **

Solving equation (1) for x,

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

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)

$\begin{array}{l}x=\frac{-2}{13}+\frac{5}{13}\left(3\right)\\ x=\frac{-2}{13}+\frac{15}{13}\\ x=1\end{array}$

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