Question

5x+4y=22 3x+5y=21 in elimination
method
$5x + 4y = 22 \\ 3x + 5y = 21$
​

Answer 1

Answer:

using substitution method

5x + 4y = 22 A

3x + 5y =21 B

from A

5x = 22- 4y

or x=(22- 4y)/5 C

substituting the value of x from C in B

3(22- 4y)/5 + 5y =21

or 3(22- 4y) + 25y =105 multiplying both sides by 5

or 66-12y+25y=105

or 13y=105-66=39

or y=39/13=3

from C

x=(22-12)/5=2

so x=2 & y=3

using elimination method

3x + 2y =10 D

x + 2y =7 E

multiplying both sides of E by 3

3x + 6y =21 F

3x + 2y =10 D

4y=11

or y=11/4

from E

x+2*11/4=7

or x=7-11/2=3/2

so x=3/2& y=11/4

ans

Answer 2

Answer:

Given Equations :

$\: \: \: \: \: \: \: \: \: { \pmb{ \sf{5x + 4y = 22...(1)}}}$

$\: \: \: \: \: \: \: \: \: { \pmb{ \sf{3x + 5y = 21...(2)}}}$

Now,

• Multiply 5 With Equation 1

• Multiply 4 with Equation 2

We Get,

$\: \: \: \: \: \: \: \: \: { \pmb{ \sf{25x + 20y = 110...(3)}}}$

$\: \: \: \: \: \: \: \: \: { \pmb{ \sf{12x + 20y = 84...(4)}}}$

Do equation (3) - equation (4) :

${ \longrightarrow{ \sf{25x - 12x + 20y - 20y = 110 - 84}}}$

By Solving We get,

${ \longrightarrow{ \sf{13x = 26}}}$

${ \longrightarrow{ \bf{ \sf{ \bf{x = 2}}}}}$

Finding Y,

${\longrightarrow{\sf{12(2) + 20y = 84}}}$

${\longrightarrow{\sf{24 + 20y = 84}}}$

${\longrightarrow{\sf{20y = 60}}}$

y = 3

Therefore,

• Value of x = 2 , y = 3