Chapter 0.1, Problem 38E

To determine

To find the pair $\left(x,y\right)$ that satisfies both the equations $3x+2y=4$ and $12x-5y=3$ simultaneously.

Answer to Problem 38E

The pair $\left(x,y\right)$ that satisfies both the equations $3x+2y=4$ and $12x-5y=3$ simultaneously:

  $\left(x,y\right)=\left(\frac{2}{3},1\right)$ .

Explanation of Solution

Given the equations:

  $3x+2y=4$ and $12x-5y=3$ .

Concept Used:

Method of substitution in the case of linear equations involving two variables:

Given a system of linear equations in two variables:

  $\left\{\begin{array}{c}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}\right.$ .

The method of substitution employs solving one of the equations, say $a_{1}x+b_{1}y=c_1$ , to find an algebraic expression for one of its variables, say $x$ .

Now, this expression of $x$ is substituted in the other equation $a_{2}x+b_{2}y=c_2$ and then solved to find the value of $y$ .

The value of $y$ is now substituted in either $a_{1}x+b_{1}y=c_1$ or $a_{2}x+b_{2}y=c_2$ to find the value of $x$ .

Now, the value of $x$ and that of $y$ are recorded to be the solution of the system of equations.

Calculation:

To find the pair $\left(x,y\right)$ that satisfies both the equations $3x+2y=4$ and $12x-5y=3$ simultaneously, it is enough the solve the system of equations $\left\{\begin{array}{c}3x+2y=4\\ 12x-5y=3\end{array}\right.$ for $x$ and $y$ .

Solve the system of equations:

Consider the first equation $3x+2y=4$ .

Isolate $y$ on the left side of the equation $3x+2y=4$ :

  $\begin{array}{c}3x+2y=4\\ 2y=4-3x\\ y=\frac{4-3x}{2}\\ y=2-\frac{3}{2}x\end{array}$ .

Substitute $y=2-\frac{3}{2}x$ in the second equation $12x-5y=3$ and solve for $x$ :

  $\begin{array}{c}12x-5\left(2-\frac{3}{2}x\right)=3\\ 12x-10+\frac{15}{2}x=3\\ 12x+\frac{15}{2}x=3+10\\ \left(12+\frac{15}{2}\right)x=13\\ x=\frac{13}{\left(12+\frac{15}{2}\right)}\\ x=\frac{2}{3}\end{array}$ .

Thus, it is found that:

  $x=\frac{2}{3}$ .

Substitute $x=\frac{2}{3}$ in the second equation $12x-5y=3$ and solve for $y$ :

  $\begin{array}{c}12x-5y=3\\ 12\left(\frac{2}{3}\right)-5y=3\\ 8-5y=3\\ -5y=3-8\\ -5y=-5\\ y=\frac{-5}{-5}\\ y=1\end{array}$ .

Thus, it is found that:

  $y=1$ .

Thus, the solution of the system of equations $\left\{\begin{array}{c}3x+2y=4\\ 12x-5y=3\end{array}\right.$ :

  $x=\frac{2}{3}$ and $y=1$ .

That is, the pair $\left(x,y\right)$ that satisfies both the equations $3x+2y=4$ and $12x-5y=3$ simultaneously:

  $\left(\frac{2}{3},1\right)$ .

Conclusion:

The pair $\left(x,y\right)$ that satisfies both the equations $3x+2y=4$ and $12x-5y=3$ simultaneously:

  $\left(x,y\right)=\left(\frac{2}{3},1\right)$ .