Using graphical method, solve the following system of equations

Q) Using graphical method, solve the following system of equations : 3x + y + 4 = 0 and 3 x – y + 2 = 0

Ans:

Step 1: Let’s try to find the intersection points on X – axis and Y – axis for each of the lines:

A. For linear equation 3 x + y + 4 = 0:

From the given equation, x = $\frac{- y - 4}{3}$

For X – axis: y = 0

∴ x = $\frac{- 0 - 4}{3} = \frac{- 4}{3}$

∴ point on X – axis: ( $\frac{- 4}{3}$ , 0)

Similarly, from the given equation, y = – 3 x – 4

For Y – axis: x = 0

∴ y = – 3 (0) – 4 = – 4

∴ point on Y – axis: (0, – 4)

B. For linear equation 3 x – y + 2 = 0:

From the given equation, x = $\frac{y - 2}{3}$

For X – axis: y = 0

∴ x = $\frac{0 - 2}{3} = \frac{- 2}{3}$

∴ point on X – axis: ( $\frac{- 2}{3}$ , 0)

Similarly, from the given equation, y = 3 x + 2

For Y – axis: x = 0

∴ y = 3 (0) + 2 = 2

∴ point on Y – axis: (0, 2)

Step 2: To represent the equations graphically, we plot the points P(0, – 4) and Q ( $\frac{- 4}{3}$ , 0) to get the line PQ.

Similarly, we plot the points R(0, 2) and S ( $\frac{- 4}{3}$ , 0) to get the line RS.

Here, the lines are intersecting each other at (- 1, – 1). hence, x = – 1 and y = – 1 lies on both the lines.

Therefore, ( – 1, – 1) is the solution for both the lines.

Check: If we add the equations, 3 x + y + 4 = 0 and 3 x – y + 2 = 0, we get 6 x + 6 = 0, hence, x = – 1.

If we deduct the equations, 3 x + y + 4 = 0 and 3 x – y + 2 = 0, we get 2 y + 2 = 0, hence, y = – 1.

Therefore above solution is correct.

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