Chapter 5, Problem 1MC

In Exercises 1–12, solve each system by the method of your choice.

1.     $\{\begin{array}{l}x=3y-7\\ 4x+3y=2\end{array}$

To determine

To solve: The system $\{\begin{array}{l}x=3y-7\\ 4x+3y=2\end{array}$ by using the substitution.

Answer to Problem 1MC

The solution set is $\left\{\left(-1,2\right)\right\}$.

Explanation of Solution

Method used:

Solving Linear Systems by Substitution:

1. Solve either of the equations for one variable in terms of the other. (If one of the equations is already in this form, you can skip this step).

2. Substitute the expression found in step 1into the other equation. This will result in an equation in one variable.

3. Solve the equation containing one variable.

4. Back-substitute the value found in step 3 into one of the original equations. Simplify and find the value of the reaming variable.

5. Check the proposed solution in both of the system`s given equations.

Calculation:

The solution of the system $\{\begin{array}{l}x=3y-7\\ 4x+3y=2\end{array}$ is as follows.

Step 1: Solve either of the equations for one variable in terms of the other:

In above system observed that the equation $x=3y-7$ is already in the form of one variable.

Step 2: Substitute the expression found in step 1 into the other equation:

Substitute $3y-7$ for $x$ in the second equation $4x+3y=2$,

$\begin{array}{l}4\left(3y-7\right)+3y=2\\ 12y-28+3y=2\\ 15y=2+28\\ 15y=30\end{array}$

From the above equation it is observed that the variable $x$ has been eliminated.

Step 3: Solve the resulting equation containing one variable:

Solve the equation $15y=30$ for $y$,

$\begin{array}{l}15y=30\\ \frac{15y}{15}=\frac{30}{15}\left[\text{Divide both sides by 15}\right]\\ y=\frac{30}{15}\\ y=2\end{array}$

Step 4: Back-substitute the obtained value into the one of the original equations:

Substitute 2 for $y$ in the one of the original equations to find $x$.

Substitute 2 for $y$ in the first equation,

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

Similarly, substitute 2 for $y$ in the second equation,

$\begin{array}{l}4x+3\left(2\right)=2\\ 4x+6=2-6\\ 4x=-4\\ x=-1\end{array}$

Here, it is cleared that the obtained values of $x$ in either case are the same.

Thus, with $x=-1$ and $y=2$, the proposed solution of the system is $\left(-1,2\right)$.

5. Check:

Check whether the proposed solution $\left(-1,2\right)$, satisfies both equations or not.

Substitute $-1$ for $x$ and $2$ for $y$ in the first equation,

$\begin{array}{l}-1=3\left(2\right)-7\\ -1=6-7\\ -1=-1\text{true statement}\end{array}$

Similarly, substitute $-1$ for $x$ and $2$ for $y$ in the second equation,

$\begin{array}{l}4\left(-1\right)+3\left(2\right)=2\\ -4+6=2\\ 2=2\text{true statement}\end{array}$

Therefore, the solution set is $\left\{\left(-1,2\right)\right\}$.