Chapter 11, Problem 1RE

For the following exercise, determine whether the ordered pair is a solution to the system of equations.

$\text{3}x-y=\text{4}$

$x+\text{4}y=-\text{3}$ and $\left(-1,1\right)$

To determine

To Check:

Whether the given ordered pair is a solution of the system.

Explanation of Solution

Given information:

$3x-y=4;x+4y=-3$with $\left(-1,1\right)$

Concept Involved:

Solving a system of equation means finding the value of (x, y) which will make both the equation TRUE.

• Here in this problem to check whether the given point is solution of system of equation, we need to substitute the value of x and y given into system of equation, and check whether it makes both the equation TRUE.

• If and only if given point (x, y) makes both the equation TRUE, graphically (x, y) will be point were the two given lines MEET and it is called a solution of system of equation.

There are different types of solution for system of equation.

Case 1: When we get intersecting lines, the solution of system of equations is UNIQUE and it is the point where the graphs intersect. The solution set in interval notation is given by $\left\{(x,y)\right\}$. The solution is described as consistent and independent. Also for intersecting lines, the slope of the system of equations will be DIFFERENT.

Case 2: When we get coinciding lines, the solution of system of equations is INFINITE because the lines meet at numerous points. The solution is represented as $\left\{x\in \Re ,y\in \Re \right\}$

The solution is described as consistent and dependent. Also for intersecting lines, the slope and y-intercept of the system of equations will be SAME.

Case 3: When we have parallel lines, the solution of the system of equations is NO SOLUTION because the lines will never meet. The solution is represented as $\left\{\right\}$or $\varnothing$. The system of equation hence described as inconsistent. Also for parallel lines, both slope and y-intercept of the equation will be DIFFERENT.

Calculation:

Description	$3x-y=4;x+4y=-3$with $\left(-1,1\right)$
Step 1: Let us label the system of equations as 1stand 2ndequation	$\{\begin{array}{l}3x-y=4\to 1^{st}equation\\ x+4y=-3\to 2^{nd}equation\end{array}$
Step 2: In an attempt of checking whether the given ordered pair $\left(-1,1\right)$ is solution to the system of equation by Substituting   $x=-1,y=1$ in 1stand check whether it is TRUE or FALSE.	$\begin{array}{l}3x-y=4\\ 3\left(-1\right)-1=4\Rightarrow -3-1=4\\ -4=4\end{array}$FALSE
Step 3: In an attempt of checking whether the given ordered pair $\left(-1,1\right)$ is solution to the system of equation by Substituting   $x=-1,y=1$ in 2ndand check whether it is TRUE or FALSE.	$\begin{array}{l}x+4y=-3\\ -1+4(1)=-3\Rightarrow -1+4=-3\\ 3=-3\end{array}$FALSE
Step 4: Since the ordered pair $\left(-1,1\right)$ makes 1stequationFALSE and 2ndequation FALSE, it is the solution of the given system of equation	$\left(-1,1\right)$is a Solution to the below system of equation $\{\begin{array}{l}3x-y=4\to 1^{st}equation\\ x+4y=-3\to 2^{nd}equation\end{array}$

Conclusion:

The ordered pair $\left(-1,1\right)$makes both 1^stand 2^ndequation FALSE; it is the not a solution of the given system of equation.