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Question: Solve the equations \[3x + 2y = 5\] and \[2x - 3y = 7\] graphically....

Solve the equations 3x+2y=53x + 2y = 5 and 2x3y=72x - 3y = 7 graphically.

Explanation

Solution

: Linear equations in two variables are straight lines when they are plotted on a graph. And, if two straight lines intersect at a point, then the point gives us the solution for the equations. We use this concept to solve this problem. We will also know about some rules that have to be followed to find the consistency and nature of solutions.

Complete step by step solution:
If we have two equations, say a1x+b1y+c1=0{a_1}x + {b_1}y + {c_1} = 0 and a2x+b2y+c2=0{a_2}x + {b_2}y + {c_2} = 0 , then these two equations are –
Consistent and have unique solution if a1a2b1b2\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} , and the two straight lines represented by these two equations are intersecting.
Consistent and have infinitely many solutions if a1a2=b1b2=c1c2\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} , and the two lines are coinciding lines.
Inconsistent and have no solution if a1a2=b1b2c1c2\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}} , and the two lines are parallel.
So, in the question, a1a2=32\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{3}{2} and b1b2=23\dfrac{{{b_1}}}{{{b_2}}} = - \dfrac{2}{3} . So, here, a1a2b1b2\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} which means, the equations have a unique solution.
To draw a graph for an equation ax+by+c=0ax + by + c = 0 , we need to follow these steps-
First of all, consider the value of xx to be zero, and calculate the value of yy by substituting x=0x = 0 in the equation, and you will get one of the points on line as (0,yat x=0)(0,{y_{{\text{at }}x = 0}}) .
Now, consider, the value of yy to be zero, and calculate the value of xx by substituting y=0y = 0 in the equation, and you will get another point on line as (xat y=0,0)({x_{{\text{at }}y = 0}},0) .
Now, join these two points, and you will get the graph for this equation.
Similarly, draw the graph for another equation also, and find the solution accordingly.
Here, in the equation 3x+2y=53x + 2y = 5 , if x=0x = 0 , we get y=52y = \dfrac{5}{2} , so one point is (0,52)\left( {0,\dfrac{5}{2}} \right) .
If y=0y = 0 , we get x=53x = \dfrac{5}{3} so, the point is (53,0)\left( {\dfrac{5}{3},0} \right) . Join these two points to get a straight line.
And in the equation, 2x3y=72x - 3y = 7 , if x=0x = 0 , we get y=73y = - \dfrac{7}{3} , so the point is (0,73)\left( {0, - \dfrac{7}{3}} \right) and if y=0y = 0 , we get x=72x = \dfrac{7}{2} , so the point is (72,0)\left( {\dfrac{7}{2},0} \right) . Join these two points to get another straight line.
These two straight lines meet at a point (2913,1113)\left( {\dfrac{{29}}{{13}}, - \dfrac{{11}}{{13}}} \right) .

Note:
A linear equation in two variables is always a straight line. Before solving any question, make sure that you check the consistency and also the nature of the solution. And be careful while taking the points. If we consider x=0x = 0 , that means that we are finding the intersection point of the given line and the Y-axis. And similarly, if we consider y=0y = 0 , that means that we are finding the intersection point of the given line and the X-axis.