Question

solve the following pair of linear equations by substitution and cross multiplication methods 8x+5y=9 and 3x+2y=4​

Answer 1

Answer:

Here's your answer

Hope it helps you.....

Answer 2

Given:-

• 8x + 5y = 9
• 3x +2y = 4

To find:-

• Solve the linear equations by substitution and cross multiplication methods.

Solutions:-

• 8x + 5y = 9 ..........(i).
• 3x +2y = 4 ...........(ii).

From equation (ii).

3x + 2y = 4

• x = (4 - 2y)/3 .......(iii).

Substitution the value of in equation (ii), we get.

8x + 5y = 9

8[(4 - 2y)/3] + 5y = 9

32 - 16y + 15y = 27

-y = 27 - 32

-y = -5

y = 5

Putting the value of y in Eq (ii), we get.

=> 3x + 2y = 4

=> 3x + 2(5) = 4

=> 3x + 10 = 4

=> 3x = 4 - 10

=> 3x = -6

=> x = -6/3

=> x = -2

Hence, x = -2 and y = 5.

Again, by cross - multiplication methods.

• 8x + 5y - 9 = 0 ..........(i).
• 3x +2y - 4 = 0 ...........(ii).

$\leadsto \: \: \frac{x}{{({5} \: \times \: {4})} \: - \: {({9} \: \times \: {-2})}} \: = \: \frac{y}{{({-9} \: \times \: {3})} \: - \: {({-5} \: \times \: {8})}} \: = \: \frac{1}{{({8} \: \times \: {2})} \: - \: {({3} \: \times \: {5})}}$

$\leadsto \: \: \frac{x}{{-20} \: - \: {(18)}} \: = \: \frac{y}{{-27} \: + \: {32}} \: = \: \frac{1}{1}$

$\leadsto \: \: \frac{x}{-2} \: = \: \frac{y}{5} \: = \: \frac{1}{1}$

$\leadsto \: \: \frac{x}{-2} \: = \: {1} \: \: \: \: \: \: and \: \: \: \: \: \: \frac{y}{5} \: = \: {1}$

=> x = -2 and y = 5

Hence, x = -2 and y = 5.