Question

How do you solve the simultaneous equations 10x−8y=7 and 3x−2y=2?

Answer 1

x=1/2, y=-1/4 is the answer.

Answer 2

10x - 8y = 7 ........ (i)

3x - 2y = 2 ....... (ii)

These are two linear equations which we can solve through Elimination method, substitution method, or cross - multiplication method.

1. Elimination method:

In this method, we've to eliminate one of the variable to get another's value.

10x - 8y = 7

3x - 2y = 2

Multiplying equation 2 by 4.

4(3x - 2y = 2)

12x - 8y = 8 ..... (iii)

Subtracting equation (i) from (iii),

12x - 8y = 8

10x - 8y = 7

(-) (+) (-)

____________

2x + 0 = 1

2x = 1

$\fbox{\bold{\mathsf{x = \frac{1}{2}}}}$

Substituting x = $\frac{1}{2}$ in equation (ii).

3x - 2y = 2

3$(\frac{1}{2})$ - 2y = 2

$\frac{3}{2} - 2y = 2$

$\frac{3 - 4y}{2} = 2$ (cross - multiplication)

3 - 4y = 4

-4y = 4 - 3

-4y = 1

$\fbox{\bold{\mathsf{y = \frac{-1}{4}}}}$

Therefore, x = $\frac{1}{2} \& y = \frac{-1}{4}$ are the solutions for the given equations.

2. Substitution method:

Here, we've to convert anyone of these equations either in x form or in y form.

10x - 8y = 7 ........ (i)

3x - 2y = 2 ....... (ii)

3x = 2 + 2y

x = $\frac{2 + 2y}{3}$.... (iii)

Substituting x = $\frac{2 + 2y}{3}$ in equation (i).

10x - 8y = 7

10$(\frac{2+2y}{3}) - 8y = 7$

$\frac{20+20y}{3} - 8y = 7$

$\frac{20+20y-24y}{3} = 7$ (cross - multiplication)

$\frac{20+(-4y)}{3} = 7$

$\frac{20-4y}{3} = 7$

20 - 4y = 21

-4y = 21 - 20

-4y = 1

$\fbox{\bold{\mathsf{y = \frac{-1}{4}}}}$

Substituting y = $\frac{-1}{4}$ in equation (ii).

3x - 2y = 2

3x - 2$(\frac{-1}{4})$ = 2

3x + $\frac{2}{4}$ = 2

3x + $\frac{\cancel{2}}{\cancel{4}}$ = 2

3x + $\frac{1}{2}$ = 2

$\frac{1+6x}{2}$ = 2

1 + 6x = 4

6x = 4 - 1

6x = 3

x = $\frac{3}{6}$

x = $\frac{\cancel{3}}{\cancel{6}}$

$\boxed{\bold{\mathsf{x = \frac{1}{2}}}}$

Therefore, x = $\frac{1}{2} \& y = \frac{-1}{4}$ are the solutions for the given equations.

3. Cross - multiplication method:

Here, we've to cross - multiply theq co-efficient of x and y and the constant term.

10x - 8y = 7 ........ (i)

3x - 2y = 2 ....... (ii) 2

10x - 8y - 7 = 0 ...... (iii)

3x - 2y - 2 = 0 ......(iv)

Comparing equation (iii) and (iv) with a1x + b1y + c1 = 0 and a2 + b2 + c2 = 0,

a1 = 10, b1 = - 8, c1 = - 7, a2 = 3, b2 = - 2, c2 = - 2

$\frac{x}{16 - 14} = \frac{y}{-20 - (-21) } = \frac{1}{-20-(-24)}$

$\frac{x}{2} = \frac{y}{(-20+21)} = \frac{1}{(-20+24)}$

$\frac{x}{2} = \frac{y}{1} = \frac{1}{4}$

$\frac{x}{2} = \frac{1}{4}$

$\frac{x}{\cancel{2}} = \frac{1}{\cancel{4}}$

$\frac{x}{1} = \frac{1}{2}$

2x = 1 (cross - multiplication)

$\boxed{\bold{\mathsf{x = \frac{1}{2}}}}$

$\frac{y}{1} = \frac{1}{4}$

4y = 1

$\boxed{\bold{\mathsf{y = \frac{1}{4}}}}$

Therefore, x = $\frac{1}{2} \& y = \frac{-1}{4}$ are the solutions for the given equations.