Chapter 11, Problem 1P

Solve a system of two linear equations two variables.

1. (a) $\begin{array}{c}5x+2y=20\\ 3x-2y=20\end{array}$
2. (b) $\begin{array}{l}x=3y\\ 2y-4x=30\end{array}$
3. (c) $\begin{array}{c}3x-5y=7\\ 6x-10y=14\end{array}$
4. (d) $\begin{array}{l}2y=5x-7\\ 10x-4y=9\end{array}$

To determine

To solve: The system of equation $5x+2y=20\text{ and 3}x-2y=12$.

Answer to Problem 1P

The solution for the system of equations $5x+2y=20\text{ and 3}x-2y=12$ is $\underset{\_}{\left(4,0\right)}$.

Explanation of Solution

Procedure used:

Solving linear systems by Addition:

(1) If necessary, rewrite both equation in the form of $Ax+By=C$.

(2) If necessary, multiply either equation or both equation by appropriate nonzero numbers so that the sum of the x-coefficient or the sum of the y-coefficients is 0.

(3) Add the equation in step 2. The sum is an equation in one variable.

(4) Solve the equation in one variable.

(5) Back-substitute the value found in step 4 either of the given equations and solve for the other variable.

(6) Check the solution in either of the original equations.

Calculation:

Consider the given equations.

$5x+2y=20$        (1)

$\text{3}x-2y=12$        (2)

By using the above procedure, solve the system of equations.

Step 1:

The equations (1) and (2) are in the form $Ax+By=C$.

Step 2:

Let eliminate the term y.

Since the coefficient of y term in both equations are opposite to each other, no need to multiply the equation.

Step 3:

Now add the equation (1) and (2) as follows.

$\begin{array}{l}5x+2y=20\\ \underset{\_}{3x-2y=12}\\ 8x+0y=32\end{array}$

Step 4:

Solve the above equation.

$\begin{array}{l}8x+0y=32\\ \frac{8x}{8}=\frac{32}{8}\left[\text{Divide by 8 on both sides}\right]\\ x=4\end{array}$

Step 5:

Now, the x-coordinate of the solution is known.

Back-substitute 4 for x in the equations (1) and obtain the value of y.

$\begin{array}{l}5\left(4\right)+2y=20\\ 20+2y-20=20-20\left[\text{Subtract 20 on both sides}\right]\\ 2y=0\\ y=0\end{array}$

Thus, the required solution is $\underset{\_}{\left(4,0\right)}$.

Step 6:

Check whether the obtained solution $\left(4,0\right)$ satisfies the given equations or not.

Replace x with 4 and y with 0 in the equation (2).

$\begin{array}{l}3x-2y=12\\ 3\left(4\right)-2\left(0\right)=12\\ 12-0=12\\ 12=12\end{array}$

Here, the ordered pair $\left(4,0\right)$ satisfies equation (1).

Hence, the results checked.

To determine

To solve: The system of equations $x=3y\text{ and }2y-4x=30$.

Answer to Problem 1P

The solution for the system of equations $x=3y\text{ and }2y-4x=30$ is $\underset{\_}{\left(-9,-3\right)}$.

Explanation of Solution

Procedure used:

Solving system of equation by Substitution method:

(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 1 into the other equation.

(3) Solve the equation containing one variable.

(4) Back-substitute the value found in step 3 into one of the original equations and find the value of the remaining variable.

(5) Check the solution by substituting in one of the system of given equations”.

Calculation:

Consider the given equation as follows.

$x=3y$        (3)

$2y-4x=30$        (4)

By using the above procedure, solve the system of equations.

Step 1:

Here the equation (3) is already in terms of y for x.

So isolation is not required.

Step 2:

Substitute $x=3y$ in the equation (4).

$\begin{array}{l}2y-4x=30\\ 2y-4\left(3y\right)=30\end{array}$

Here, the variable x is eliminated.

Step 3:

Solve the above equation.

$\begin{array}{l}2y-4\left(3y\right)=30\\ 2y-12y=30\left(\text{Apply the distributive property}\right)\\ -10y=30\left(\text{Combine like terms}\right)\\ \frac{-10y}{-10}=\frac{30}{-10}\left(\text{Divide by }-10\text{ on both sides}\right)\\ y=-3\end{array}$

Step 4:

Now, the y-coordinate of the solution is known.

Back-substitute –3 for y in the equations (3) and obtain the value of x.

$\begin{array}{c}x=3y\\ =3\left(-3\right)\\ =-9\end{array}$

Thus, the required solution is $\underset{\_}{\left(-9,-3\right)}$.

Step 5:

Check whether the obtained solution $\left(-9,-3\right)$ satisfies the given equations or not.

Replace x with –9 and y with –3 in the equation (4).

$\begin{array}{l}2\left(-3\right)-4\left(-9\right)=30\\ -6+36=30\\ 30=30\end{array}$

Here, the ordered pair $\left(-9,-3\right)$ satisfies equation (4).

To determine

To solve: The system of equation $3x-5y=7\text{ and }6x-10y=14$.

Answer to Problem 1P

The system of equation $3x-5y=7\text{ and }6x-10y=14$ has $\underset{\_}{\text{infinetly many solutions}}$.

Explanation of Solution

Definition used:

Consistent:

In solving a system of equations, if the equations produces a true statement of $0=0$, then the equation is said to be consistent.

The consistent system of equations has infinitely many solutions.

Calculation:

Consider the given equations.

$3x-5y=7$        (5)

$6x-10y=14$        (6)

By using the above procedure, solve the system of equations.

Step 1:

Let eliminate the term y.

Since the coefficients of y term are –5 and –10, multiply the equation (5) by –2.

That is, $-6x+10y=-14$        (7)

Step 3:

Now add the equation (6) and (7) as follows.

$\begin{array}{l}6x-10y=14\\ \underset{\_}{-6x+10y=-14}\\ \text{ 0}x+0y=0\end{array}$

Step 4:

Solve the above equation.

$\begin{array}{l}\text{0}y+0x=0\\ 0=0\end{array}$

Here $0=0$ is a true statement.

Since the system of equation has produced a true statement, by above definition the given system is consistent.

Therefore, the system of equation $3x-5y=7\text{ and }6x-10y=14$ has $\underset{\_}{\text{infinetly many solutions}}$.

To determine

To solve: The system of equation $2y=5x-7\text{ and }10x-4y=9$.

Answer to Problem 1P

The system of equation $2y=5x-7\text{ and }10x-4y=9$ has $\underset{\_}{\text{no solution}}$.

Explanation of Solution

Definition used:

Inconsistent:

In solving a system of equations and it produce a false statement, then the equation is said to be inconsistent.

The inconsistent system of equations has no solution.

Calculation:

Consider the given equations.

$2y=5x-7$        (8)

$10x-4y=9$        (9)

By using the above procedure, solve the system of equations.

Step 1:

Let isolate the term y from the equation (8).

$\begin{array}{l}2y=5x-7\\ \frac{2y}{2}=\frac{5x-7}{2}\\ y=\frac{5x-7}{2}\end{array}$

Step 3:

Substitute $y=\frac{5x-7}{2}$ in the equation (9).

$\begin{array}{l}10x-4y=9\\ 10x-4\left(\frac{5x-7}{2}\right)=9\end{array}$

Here, the variable x is eliminated.

Step 3:

Solve the above equation.

$\begin{array}{l}10x-4\left(\frac{5x-7}{2}\right)=9\\ 10x-10x-14=9\left(\text{Apply the distributive property}\right)\\ 0x=9+14\left(\text{Combine like terms}\right)\\ 0=25\end{array}$

Here $0=25$ is a false statement.

Since the system of equation has produced a false statement, by above definition the given system is inconsistent.

Therefore, the system of equation $2y=5x-7\text{ and }10x-4y=9$ has $\underset{\_}{\text{no solution}}$.