Question

Solve the following equations by Cramer’s method.

i. 6x - 3y = -10; 3x + 5y - 8 = 0

ii. 4m - 2n = -4; 4m + 3n = 16

iii. 3x - 2y = \(\frac{5}{2};\) \(\frac{1}{3}x + 3y = -\frac{4}{3}\)

iv. 7x + 3y = 15; 12y - 5x = 39

v. \(\frac{x + y - 8}{2}\) = \(\frac{x + 2y - 14}{3}\) = \(\frac{3x - y}{4}\)

Answer 1

i. The given simultaneous equations are

6x – 3y = -10 …(i) 

3x + 5y – 8 = 0 

∴ 3x + 5y = 8 …(ii)

Equations (i) and (ii) are in ax + by = c form.

Comparing the given equations with a₁x + b₁y = c₁

and a₂x + b₂y = c₂, we get

a₁ = 6, b₁ = -3, c₁ = 10 and

ii. The given simultaneous equations are 

4m – 2n = -4 …(i) 

4m + 3n = 16 …(ii)

Equations (i) and (ii) are in am + bn = c form. Comparing the given equations with a₁m + b₁n = c₁

∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are 

7x + 3y = 15 …(i) 

12y – 5x = 39 

i.e. -5x + 12y = 39 …(ii) 

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with

v. The given simultaneous equations are

[Dividing both sides by 2]

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.