Chapter 0.1, Problem 37E

To determine

To find the pair $\left(x,y\right)$ that satisfies both the equations $4x-y=4$ and $14x+3y=1$ simultaneously.

Answer to Problem 37E

The pair $\left(x,y\right)$ that satisfies both the equations $4x-y=4$ and $14x+3y=1$ simultaneously:

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

Explanation of Solution

Given the equations:

  $4x-y=4$ and $14x+3y=1$ .

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 $4x-y=4$ and $14x+3y=1$ simultaneously, it is enough the solve the system of equations $\left\{\begin{array}{c}4x-y=4\\ 14x+3y=1\end{array}\right.$ for $x$ and $y$ .

Solve the system of equations:

Consider the first equation $4x-y=4$ .

Isolate $y$ on the left side of the equation $4x-y=4$ :

  $\begin{array}{c}4x-y=4\\ -y=4-4x\\ y=4x-4\end{array}$ .

Substitute $y=4x-4$ in the second equation $14x+3y=1$ and solve for $x$ :

  $\begin{array}{c}14x+3\left(4x-4\right)=1\\ 14x+12x-12=1\\ 14x+12x=1+12\\ 26x=13\\ x=\frac{13}{26}\\ x=\frac{1}{2}\end{array}$ .

Thus, it is found that:

  $x=\frac{1}{2}$ .

Substitute $x=\frac{1}{2}$ in the second equation $14x+3y=1$ and solve for $y$ :

  $\begin{array}{c}14x+3y=1\\ 14\left(\frac{1}{2}\right)+3y=1\\ 7+3y=1\\ 3y=1-7\\ 3y=-6\\ y=\frac{-6}{3}\\ y=-2\end{array}$ .

Thus, it is found that:

  $y=-2$ .

Thus, the solution of the system of equations $\left\{\begin{array}{c}4x-y=4\\ 14x+3y=1\end{array}\right.$ :

  $x=\frac{1}{2}$ and $y=-2$ .

That is, the pair $\left(x,y\right)$ that satisfies both the equations $4x-y=4$ and $14x+3y=1$ simultaneously:

  $\left(\frac{1}{2},-2\right)$ .

Conclusion:

The pair $\left(x,y\right)$ that satisfies both the equations $4x-y=4$ and $14x+3y=1$ simultaneously:

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