# Examine which of the following pair of values of x and y is a solution of equation 4x — 3y + 24 = 0.(i) x = 0, y = 8 (ii) x = — 6, y = 0(iii) x = 1, y = — 2 (iv) x = – 3,y = 4(v) x = 1, y = — 2 (vi) x = — 4, y = 2

Given equation is

i) Justification

On substituting x = 0, y = 8 in LHS of given equation, we get

LHS = 4(0) – 3(8) + 24 = 0 – 24 + 24 = 0 = RHS

Hence, x = 0, y = 8 is a solution of the equation 4x – 3y + 24 = 0

ii) Justification

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

LHS = 4( – 6) – 3(0) + 24 = – 24 + 24 = 0 = RHS

Hence, x = – 6, y = 0 is a solution of the equation 4x – 3y + 24 = 0

iii) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

iv) Justification

On substituting x = – 3, y = 4 in LHS of given equation, we get

LHS = 4( – 3) – 3(4) + 24 = – 12 – 12 + 24 = 0 = RHS

Hence, x = – 3, y = 4 is a solution of the equation 4x – 3y + 24 = 0

v) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

vi) Justification

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

LHS = 4( – 4) – 3(2) + 24 = – 16 – 6 + 24 = – 22 + 24 = 2 ≠ RHS

Hence, x = – 4, y = 2 is not a solution of the equation 4x – 3y + 24 = 0

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