Question

Solve the following systems of linear equations graphically

(a) If the system has a unique solution, check your answer by substituting in the equation.

(b) If the system is inconsistent or dependent, state which of the two it is?

$$\begin{gathered} 4x-y+3=0 \\ x=y \end{gathered}$$

Answer 1

Step 1: Given data:

Given equation are

$$\begin{gathered} 4x-y+3=0-----\left(1\right) \\ x=y-----\left(2\right) \end{gathered}$$

Step 2: Find the points corresponding to equation(1)

Writing equation(1) in terms of x as

$y=4x+3$

substituting different values for x to get the values of y as follows

$$\begin{gathered} x=0\Rightarrow y=3 \\ x=-1\Rightarrow y=-4+3=-1 \\ x=-2\Rightarrow y=-5 \\ therefore(0,3),(-1,-1)and(-2,-5)arepoints \end{gathered}$$

$x$	$0$	$-1$	$-2$
$y$	$3$	$-1$	$-5$

Step 3: Find the points corresponding to equation(2)

$y=x$

substituting different values of x in (2),we get different values of y as follows

$$\begin{gathered} x=0\Rightarrow y=0 \\ x=1\Rightarrow y=1 \\ x=-1\Rightarrow y=-1 \\ ie,(0,0),(1,1)and(-1,-1)arepoints. \end{gathered}$$

$x$	$0$	$1$	$-1$
$y$	$0$	$1$	$-1$

step 4: plotting the points on a graph

From the graph, we can see that the lines are intersecting at the point$(-1,-1)$

i.e. the pair of equation have a unique solution

Step 5: Cheking for point $(-1,-1)$:

$$\begin{gathered} 4x-y+3=0 \\ putting,x=-1andy=-1 \\ \Rightarrow 4\times (-1)-(-1)+3=0 \\ \Rightarrow -4+1+3=0 \\ \Rightarrow -3+3=0 \\ \Rightarrow 0=0 \end{gathered}$$

$$\begin{gathered} x=y \\ putting,x=-1andy=-1 \\ \Rightarrow -1=-1 \\ \Rightarrow -1+1=0 \\ \Rightarrow 0=0 \end{gathered}$$

Hence, the system of equations is consistent.