Question

Solve graphically each of the following systems of linear equations. Also, find the coordinates of the points where the lines meet the axis of x in each system:
$
  (i)2x + y = 6 \\
  x - 2y = - 2 \\
  (ii)2x - y = 2 \\
  4x - y = 8 \\
  (iii)x + 2y = 5 \\
  2x - 3y = - 4 \\
  (iv)2x + 3y = 8 \\
  x - 2y = - 3 \\
 $

Answer 1

Hint: Here, we will proceed by putting x = 0 in the given linear equations in order to find the corresponding value of y and then by putting y = 0 in the given linear equations to find the corresponding value of x. This will give us the passing points of these lines.

Complete step-by-step answer:

$(i)$ Given linear equations are $2x + y = 6{\text{ }} \to {\text{(1)}}$ and $x - 2y = - 2{\text{ }} \to {\text{(2)}}$
Put x = 0 in equation (1), we get
$
   \Rightarrow \left( {2 \times 0} \right) + y = 6 \\
   \Rightarrow 0 + y = 6 \\
   \Rightarrow y = 6 \\
 $
Put y = 0 in equation (1), we get
$
   \Rightarrow 2x + 0 = 6 \\
   \Rightarrow 2x = 6 \\
   \Rightarrow x = \dfrac{6}{2} = 3 \\
 $
Therefore, the straight line whose linear equation is $2x + y = 6$ has two passing points as A(0,6) and B(3,0). Now, with the help of these two points this straight line can be drawn as shown in figure 1.
Put x = 0 in equation (2), we get
$
   \Rightarrow 0 - 2y = - 2 \\
   \Rightarrow - 2y = - 2 \\
   \Rightarrow y = \dfrac{{ - 2}}{{ - 2}} = 1 \\
 $
Put y = 0 in equation (2), we get
$
   \Rightarrow x - \left( {2 \times 0} \right) = - 2 \\
   \Rightarrow x - 0 = - 2 \\
   \Rightarrow x = - 2 \\
 $
Therefore, the straight line whose linear equation is $x - 2y = - 2$ has two passing points as C(0,1) and D(-2,0). Now, with the help of these two points this straight line can be drawn as shown in figure 1.
Equation (2) can be rewritten as
$ \Rightarrow x = 2y - 2{\text{ }} \to {\text{(3)}}$
Put the value of x from equation (3) in equation (1), we have
$
   \Rightarrow 2\left( {2y - 2} \right) + y = 6 \\
   \Rightarrow 4y - 4 + y = 6 \\
   \Rightarrow 5y = 6 + 4 = 10 \\
   \Rightarrow y = \dfrac{{10}}{5} = 2 \\
 $
Put y = 2 in equation (3), we get
$
   \Rightarrow x = \left( {2 \times 2} \right) - 2 \\
   \Rightarrow x = 4 - 2 \\
   \Rightarrow x = 2 \\
 $
Therefore, the coordinates of the point where the two straight lines having linear equations as $2x + y = 6$ and $x - 2y = - 2$ meets is P(2,2).

$(ii)$ Given linear equations are $2x - y = 2{\text{ }} \to {\text{(1)}}$ and $4x - y = 8{\text{ }} \to {\text{(2)}}$
Put x = 0 in equation (1), we get
$
   \Rightarrow \left( {2 \times 0} \right) - y = 2 \\
   \Rightarrow 0 - y = 2 \\
   \Rightarrow y = - 2 \\
 $
Put y = 0 in equation (1), we get
$
   \Rightarrow 2x - 0 = 2 \\
   \Rightarrow 2x = 2 \\
   \Rightarrow x = \dfrac{2}{2} = 1 \\
 $
Therefore, the straight line whose linear equation is $2x - y = 2$ has two passing points as A(0,-2) and B(1,0). Now, with the help of these two points this straight line can be drawn as shown in figure 2.
Put x = 0 in equation (2), we get
\[
   \Rightarrow \left( {4 \times 0} \right) - y = 8 \\
   \Rightarrow 0 - y = 8 \\
   \Rightarrow y = - 8 \\
 \]
Put y = 0 in equation (2), we get
$
   \Rightarrow 4x - 0 = 8 \\
   \Rightarrow 4x = 8 \\
   \Rightarrow x = \dfrac{8}{4} = 2 \\
 $
Therefore, the straight line whose linear equation is $4x - y = 8$ has two passing points as C(0,-8) and D(2,0). Now, with the help of these two points this straight line can be drawn as shown in figure 2.
Equation (2) can be rewritten as
$ \Rightarrow y = 4x - 8{\text{ }} \to {\text{(3)}}$
Put the value of y from equation (3) in equation (1), we have
$
   \Rightarrow 2x - \left( {4x - 8} \right) = 2 \\
   \Rightarrow 2x - 4x + 8 = 2 \\
   \Rightarrow 2x = 6 \\
   \Rightarrow x = 3 \\
 $
Put x = 3 in equation (3), we get
$
   \Rightarrow y = \left( {4 \times 3} \right) - 8 \\
   \Rightarrow y = 12 - 8 = 4 \\
 $
Therefore, the coordinates of the point where the two straight lines have linear equations as $2x - y = 2$ and $4x - y = 8$ meets P(3,4).

$(iii)$ Given linear equations are \[x + 2y = 5{\text{ }} \to {\text{(1)}}\] and $2x - 3y = - 4{\text{ }} \to {\text{(2)}}$
Put x = 0 in equation (1), we get
\[
   \Rightarrow 0 + 2y = 5 \\
   \Rightarrow 2y = 5 \\
   \Rightarrow y = \dfrac{5}{2} \\
 \]
Put y = 0 in equation (1), we get
\[
   \Rightarrow x + \left( {2 \times 0} \right) = 5 \\
   \Rightarrow x + 0 = 5 \\
   \Rightarrow x = 5 \\
 \]
Therefore, the straight line whose linear equation is \[x + 2y = 5\] has two passing points as A(0,\[\dfrac{5}{2}\]) and B(5,0). Now, with the help of these two points this straight line can be drawn as shown in figure 3.
Put x = 0 in equation (2), we get
\[
   \Rightarrow \left( {2 \times 0} \right) - 3y = - 4 \\
   \Rightarrow 0 - 3y = - 4 \\
   \Rightarrow y = \dfrac{{ - 4}}{{ - 3}} \\
   \Rightarrow y = \dfrac{4}{3} \\
 \]
Put y = 0 in equation (2), we get
$
   \Rightarrow 2x - \left( {3 \times 0} \right) = - 4 \\
   \Rightarrow 2x - 0 = - 4 \\
   \Rightarrow x = \dfrac{{ - 4}}{2} \\
   \Rightarrow x = - 2 \\
 $
Therefore, the straight line whose linear equation is \[2x - 3y = - 4\] have two passing points as C(0,$\dfrac{4}{3}$) and D(-2,0). Now, with the help of these two points this straight line can be drawn as shown in figure 3.
Equation (1) can be rewritten as
$ \Rightarrow x = 5 - 2y{\text{ }} \to {\text{(3)}}$
Put the value of x from equation (3) in equation (2), we have
$
   \Rightarrow 2\left( {5 - 2y} \right) - 3y = - 4 \\
   \Rightarrow 10 - 4y - 3y = - 4 \\
   \Rightarrow 10 - 7y = - 4 \\
   \Rightarrow 7y = 14 \\
   \Rightarrow y = 2 \\
 $
Put y = 2 in equation (3), we get
$
   \Rightarrow x = 5 - \left( {2 \times 2} \right) \\
   \Rightarrow x = 5 - 4 \\
   \Rightarrow x = 1 \\
 $
Therefore, the coordinates of the point where the two straight lines having linear equations as \[x + 2y = 5\] and \[2x - 3y = - 4\] meet is P(1,2).

$(iv)$ Given linear equations are \[2x + 3y = 8{\text{ }} \to {\text{(1)}}\] and $x - 2y = - 3{\text{ }} \to {\text{(2)}}$
Put x = 0 in equation (1), we get
\[
   \Rightarrow 2 \times 0 + 3y = 8 \\
   \Rightarrow 3y = 8 \\
   \Rightarrow y = \dfrac{8}{3} \\
 \]
Put y = 0 in equation (1), we get
\[
   \Rightarrow 2x + \left( {3 \times 0} \right) = 8 \\
   \Rightarrow 2x + 0 = 8 \\
   \Rightarrow x = 4 \\
 \]
Therefore, the straight line whose linear equation is \[2x + 3y = 8\] has two passing points as A(0,\[\dfrac{8}{3}\]) and B(4,0). Now, with the help of these two points this straight line can be drawn as shown in figure 4.
Put x = 0 in equation (2), we get
\[
   \Rightarrow 0 - 2y = - 3 \\
   \Rightarrow - 2y = - 3 \\
   \Rightarrow y = \dfrac{{ - 3}}{{ - 2}} \\
   \Rightarrow y = \dfrac{3}{2} \\
 \]
Put y = 0 in equation (2), we get
$
   \Rightarrow x - \left( {2 \times 0} \right) = - 3 \\
   \Rightarrow x - 0 = - 3 \\
   \Rightarrow x = - 3 \\
 $
Therefore, the straight line whose linear equation is \[x - 2y = - 3\] have two passing points as C(0,\[\dfrac{3}{2}\]) and D(-3,0). Now, with the help of these two points this straight line can be drawn as shown in figure 4.
Equation (2) can be rewritten as
$ \Rightarrow x = 2y - 3{\text{ }} \to {\text{(3)}}$
Put the value of x from equation (3) in equation (1), we have
$
   \Rightarrow 2\left( {2y - 3} \right) + 3y = 8 \\
   \Rightarrow 4y - 6 + 3y = 8 \\
   \Rightarrow 7y - 6 = 8 \\
   \Rightarrow 7y = 14 \\
   \Rightarrow y = 2 \\
 $
Put y = 2 in equation (3), we get
$
   \Rightarrow x = \left( {2 \times 2} \right) - 3 \\
   \Rightarrow x = 4 - 3 \\
   \Rightarrow x = 1 \\
 $
Therefore, the coordinates of the point where the two straight lines having linear equations as \[2x + 3y = 8\] and \[x - 2y = - 3\] meet is P(1,2).

Note: In this particular problem, we have used the substitution method to solve for the coordinates of the meeting point between any two given lines. Here, we can also use elimination methods. Since, the meeting point of two straight lines will be satisfying the linear equations of both these lines, that’s why we have found the solution for these two lines.