Examine which of the following points lie on the graph of the linear equation 5x — 3y + 30 = 0.

(i) A (— 6, 0) (ii) B (0, 10)

(iii) C (3, — 5) (iv) D (4, 2)

(v) E (— 9, 5) (vi) F (— 3, 5)

(vii) G (— 9, — 5)

The given equation is 5x – 3y + 30 = 0

(i) Given A ( – 6,0). Here x = – 6 and y = 0

On substituting x = – 6, y = 0 in LHS of given equation, we get

LHS = 5( – 6) – 3(0) + 30 = – 30 + 30 = 0 = RHS

So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.

Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.

(ii) Given B (0,10). Here x = 0 and y = 10

On substituting x = 0, y = 10 in LHS of given equation, we get

LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS

So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0

Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.

(iii) Given C (3, – 5). Here x = 3 and y = – 5

On substituting x = 3, y = – 5 in LHS of given equation, we get

LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS

So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(iv) Given D (4,2). Here x = 4 and y = 2

On substituting x = 4, y = 2 in LHS of given equation, we get

LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS

So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(v) Given E ( – 9,5). Here x = – 9 and y = 5

On substituting x = – 9, y = 5 in LHS of given equation, we get

LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS

So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(vi) Given F ( – 3,5). Here x = – 3 and y = 5

On substituting x = – 3, y = 5 in LHS of given equation, we get

LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS

So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.

(vii) Given G ( – 9, – 5). Here x = 3 and y = – 5

On substituting x = – 9, y = – 5 in LHS of given equation, we get

LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS

So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.

Or Graphically

Here, we can see through the graph also that Point A, B, F and G lie on the graph of the linear equation 5x – 3y + 30 = 0