Solve the following system of linear equations

Q) Solve the following system of linear equations graphically:

x + 2 y = 3, 2 x – 3 y + 8 = 0

Ans:

Step 1: To plot the equations, let’s first find out the coordinates of points lying on these lines:

For line: x + 2 y = 3,, we calculate coordinates of various points:

at X = 0, 0 + 2 y = 3 ∴ y = $\frac{3}{2}$

at X = 1, 1 + 2 y = 3 ∴ y = 1

at X = 2, 2 + 2 y = 3 ∴ y = $\frac{1}{2}$

at X = 3, 3 + 2 y = 3 ∴ y = 0

Hence, we get the following table:

$\begin{tabular}{ |c|c|c|c| } \hline 0 & 1 & 2 & 3 \\ \hline 3/2 & 1 & 1/2 & 0 \\ \hline \end{tabular}$

Similarly for line: 2 x – 3 y + 8 = 0, we calculate coordinates of various points:

at X = 0, 2 (0) – 3 y + 8 = 0 ∴ y = $\frac{8}{3}$

at X = 1, 2 (1) – 3 y + 8 = 0 ∴ y = $\frac{10}{3}$

at Y = 0, 2 x – 3 (0) + 8 = 0 ∴ x = – 4

at Y = 1, 2 x – 3 (1) + 8 = 0 ∴ x = $\frac{ - 5}{2}$

Hence, we get the following table:

$\begin{tabular}{ |c|c|c|c| } \hline 0 & 1 & - 4 & - 5/2 \\ \hline 8/3 & 10/3 & 0 & 1 \\ \hline \end{tabular}$

Step 2: Now let’s plot both of these lines connecting each of the points:

From the diagram, we can see that the lines intersect each other at point (-1, 2)

Therefore, the solution of the lines is (- 1, 2).

Check: at (-1, 2), LHS value of (x + 2y = 3) is – 1 + 2 (2) = 3 = RHS

at (-1, 2), LHS value of (2 x – 3 y + 8 = 0) is 2 (- 1) – 3 (2) + 8 = – 2 – 6 + 8 = 0 = RHS

Since RHS values are same as LHS for both equations at (-1, 2), hence our answer is correct.

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