
Solve graphically the following systems of linear equations. Also find the coordinates of the points where the line meets the axis of y.
$
3x + 2y = 12 \\
5x - 2y = 4 \\
$
Answer
515.4k+ views
Hint:- In this question first we have to find at least two points on given lines to plot their lines on a graph. After plotting lines on graphs using obtained points, we have to find the point where these two lines intersect because it is the solution of a given system of linear equations.
Complete step-by-step answer:
We have, two lines of equations:
$
3x + 2y = 12 {\text{ eq}}{\text{.1 }} \\
5x - 2y = 4 {\text{ eq}}{\text{.2}} \\
$
Now, we have to plot given equations on a graph ; for this we require at least two points which lie on that line.
So, consider eq.1
$3x + 2y = 12{\text{ }}$
Now, put x=0 in eq.1, we get
$
\Rightarrow 2y = 12 \\
\Rightarrow y = 6 \\
$
Therefore, (0,6) lies on above line $3x + 2y = 12$.
Now, put y=0 in eq.1, we get
$
\Rightarrow 3x = 12 \\
\Rightarrow x = 4 \\
$
Therefore, (4,0) lies on line $3x + 2y = 12$.
Now put x=2 in eq.1, we get
$
\Rightarrow 3 \times 2 + 2y = 12 \\
\Rightarrow 2y = 6 \\
\Rightarrow y = 3 \\
$
Therefore, (2,3) lies on line $3x + 2y = 12$.
Now, consider eq.2
$5x - 2y = 4$
Now, put x=0 in eq.2, we get
$
\Rightarrow 0 - 2y = 4 \\
\Rightarrow y = - 2 \\
$
Therefore, (0,-2) lies on above line $5x - 2y = 4$.
Now, put y=0 in eq.2, we get
$
\Rightarrow 5x - 0 = 4 \\
\Rightarrow x = \left( {\dfrac{4}{5}} \right) \\
$
Therefore, $\left( {\dfrac{4}{5},0} \right)$ lies on line $5x - 2y = 4$.
Now put x=2 in eq.2, we get
$
\Rightarrow 5 \times 2 - 2y = 4 \\
\Rightarrow 2y = 6 \\
\Rightarrow y = 3 \\
$
Therefore, (2,3) lies on line $5x - 2y = 4$.
Now, plot given lines on a graph using obtained points.
Hence, from the graph we can see that x=2 and y=3 is the solution of the given system of equations.
We can also observe that line $3x + 2y = 12$ meet the y-axis at point (0,6).
And the line $5x - 2y = 4$ meet the y-axis at point (0,-2).
Note:- Whenever you get this type of question the key concept to solve this is to learn the concept of a system of linear equations containing two equations. The solution of such a system is the ordered pair that is a solution to both the equations of the system. Graphically, the solution is the intersection point where both lines of the system intersect.
Complete step-by-step answer:
We have, two lines of equations:
$
3x + 2y = 12 {\text{ eq}}{\text{.1 }} \\
5x - 2y = 4 {\text{ eq}}{\text{.2}} \\
$
Now, we have to plot given equations on a graph ; for this we require at least two points which lie on that line.
So, consider eq.1
$3x + 2y = 12{\text{ }}$
Now, put x=0 in eq.1, we get
$
\Rightarrow 2y = 12 \\
\Rightarrow y = 6 \\
$
Therefore, (0,6) lies on above line $3x + 2y = 12$.
Now, put y=0 in eq.1, we get
$
\Rightarrow 3x = 12 \\
\Rightarrow x = 4 \\
$
Therefore, (4,0) lies on line $3x + 2y = 12$.
Now put x=2 in eq.1, we get
$
\Rightarrow 3 \times 2 + 2y = 12 \\
\Rightarrow 2y = 6 \\
\Rightarrow y = 3 \\
$
Therefore, (2,3) lies on line $3x + 2y = 12$.
Now, consider eq.2
$5x - 2y = 4$
Now, put x=0 in eq.2, we get
$
\Rightarrow 0 - 2y = 4 \\
\Rightarrow y = - 2 \\
$
Therefore, (0,-2) lies on above line $5x - 2y = 4$.
Now, put y=0 in eq.2, we get
$
\Rightarrow 5x - 0 = 4 \\
\Rightarrow x = \left( {\dfrac{4}{5}} \right) \\
$
Therefore, $\left( {\dfrac{4}{5},0} \right)$ lies on line $5x - 2y = 4$.
Now put x=2 in eq.2, we get
$
\Rightarrow 5 \times 2 - 2y = 4 \\
\Rightarrow 2y = 6 \\
\Rightarrow y = 3 \\
$
Therefore, (2,3) lies on line $5x - 2y = 4$.
Now, plot given lines on a graph using obtained points.

Hence, from the graph we can see that x=2 and y=3 is the solution of the given system of equations.
We can also observe that line $3x + 2y = 12$ meet the y-axis at point (0,6).
And the line $5x - 2y = 4$ meet the y-axis at point (0,-2).
Note:- Whenever you get this type of question the key concept to solve this is to learn the concept of a system of linear equations containing two equations. The solution of such a system is the ordered pair that is a solution to both the equations of the system. Graphically, the solution is the intersection point where both lines of the system intersect.
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