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Questions
Solve the system of equations graphically:
2x + 3y + 5 = 0, 3x – 2y – 12 = 0
Solve the following system of equations graphically:
2x + 3y + 5 = 0, 3x – 2y – 12 = 0
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Solution
From the first equation, write y in terms of x
`y = -((5 + 2x)/3)` ...(i)
Substitute different values of x in (i) to get different values of y
For x = –1, y = `-(5 - 2)/3 = -1`
For x = 2, y = `-(5 + 4)/3 = -3`
For x = 5, y = `-(8 + 10)/3 = -5`
Thus, the table for the first equation (2x + 3y + 5 = 0) is
| x | –1 | 2 | 5 |
| y | –1 | –3 | –5 |
Now, plot the points A(–1, –1), B(2, –3) and C(5, –5) on a graph paper and join them to get the graph of 2x + 3y + 5 = 0.
From the second equation, write y in terms of x
`y = (3x - 12)/2` ...(ii)
Now, substitute different values of x in (ii) to get different values of y
For x = 0, y = `(0 - 12)/2 = -6`
For x = 2, y = `(6 - 12)/2 = -3 `
For x = 4, y = `(12 - 12)/2 = 0`
So, the table for the second equation (3x – 2y – 12 = 0) is
| x | 0 | 2 | 4 |
| y | –6 | –3 | 0 |
Now, plot the points D(0, –6), E(2, –3) and F(4, 0) on the same graph paper and join D, E and F to get the graph of 3x – 2y – 12 = 0.

From the graph it is clear that, the given lines intersect at (2, –3).
Hence, the solution of the given system of equation is (2, –3).
