Chapter 5.4, Problem 6Q

To determine

ToSolve:Thegiven below system of linear equations, and check the solutions.

  $3x-5y=13$

  $x+4y=10$

Answer to Problem 6Q

The solution of the given system of linear equations is $\left(x,y\right)=\left(6,1\right)$

Explanation of Solution

Given:

The system of linear equations

  $3x-5y=13$

  $x+4y=10$

Concept Used:

In solving a system of linear equation by substitution method. The value of one of the unknown from any of the equation is substituted into another equation and the value of second unknown is calculated.

The calculated value of second unknown is then substituted into any of the equation to find the value of the first unknown.

Calculation:

Given the system of linear equations

  $3x-5y=13\text{ - - - - - }\left(1\right)$

  $x+4y=10\text{ - - - - - }\left(2\right)$

Using the Substitution method

From equation $\left(1\right)$

  $\begin{array}{c}3x-5y=13\\ 3x-5y+5y=13+5y\text{ Add 5}y\text{ to both sides}\\ 3x=13+5y\\ \frac{3x}{3}=\frac{13}{3}+\frac{5y}{3}\text{ Divide both sides by 3}\end{array}$

  $x=\frac{13}{3}+\frac{5y}{3}$

Substituting the value of $x$ from equation $\left(1\right)$ into the equation $\left(2\right)$ , we have

  $\begin{array}{c}x+4y=10\\ \frac{13}{3}+\frac{5y}{3}+4y=10\text{ Substitue }\frac{13}{3}+\frac{5y}{3}\text{ for }x\\ \frac{13}{3}+\frac{5y}{3}+\frac{12y}{3}=\frac{30}{3}\\ \frac{13}{3}+\frac{5y}{3}+\frac{12y}{3}-\frac{13}{3}=\frac{30}{3}-\frac{13}{3}\text{ Subtract }\frac{13}{3}\text{ from both sides}\end{array}$

  $\begin{array}{c}\frac{17y}{3}=\frac{17}{3}\\ \left(\frac{17y}{3}\right)\left(\frac{3}{17}\right)=\left(\frac{17}{3}\right)\left(\frac{3}{17}\right)\text{ Multiply both sides by }\frac{3}{17}\\ y=1\end{array}$

Substituting $y=1$ into equation $\left(1\right)$

  $\begin{array}{c}3x-5y=13\\ 3x-5\left(1\right)=13\text{ Substitute }y=1\\ 3x-5=13\\ 3x-5+5=13+5\text{ Add 5 to both sides}\end{array}$

  $\begin{array}{c}3x=18\\ \frac{3x}{3}=\frac{18}{3}\text{ Divide both sides by 3}\\ x=6\end{array}$

Therefore, the solution of the given system of linear equations is $\left(x,y\right)=\left(6,1\right)$

Checking

Substituting the solution $\left(x,y\right)=\left(6,1\right)$ in left side of the first equation $3x-5y=13$

  $\begin{array}{c}3x-5y=3\left(6\right)-5\left(1\right)\text{ Substitute }\left(x,y\right)=\left(6,1\right)\\ =18-5\\ =13\text{ Equal to right hand side of the equation}\end{array}$

Substituting the solution $\left(x,y\right)=\left(6,1\right)$ in left side of the second equation $x+4y=10$

  $\begin{array}{c}x+4y=6+4\left(1\right)\text{ Substitute }\left(x,y\right)=\left(6,1\right)\\ =6+4\\ =10\text{ Equal to right hand side of the equation}\end{array}$

Hence, the solution is $\left(x,y\right)=\left(6,1\right)$ verified.

Conclusion:

The solution of the given system of linear equations is $\left(x,y\right)=\left(6,1\right)$