Question

using both subsititution method and elimination method solve 3x+2y=13 and 5x-3y=9​

Answer 1

AnsWer :

x= 3 and y = 2.

Given :

• General equation a1x + b1y +c= 0. and a2x + b2y + c= 0.
• We have equation, 3x+ 2y = 13. and 5x -3y = 9.

We have Method.

• Elimination method.
• Substitution method.
• Cross Multiplication method.

Solution :

Let's try Elimination method

We have equation,

$\tt3x + 2y = 13. - - - (1)$

$\tt5x - 3y = 9. - - - (2)$

In equation 1 Multiply 3 and Equation 2 Multiply by 2, we get.

$\begin{aligned} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 9x +&6y =39 \\ 10x - &6y =18. \\ \end{aligned}$

____________________________

$\begin{aligned} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 19x = 57 \end{aligned}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 3.$

Now, Putting the value x = 3 in equation (1), we get.

$\implies3x + 2y = 13. \\ \implies9 + 2y = 13. \\ \implies2y = 4. \\ \implies y = 2.$

$\rule{200}3$

Let's try Substitution method.

We have equation,

$\tt3x + 2y = 13. - - - (1)$

$\tt5x - 3y = 9. - - - (2)$

Taking equation 3,

$\tt \implies5x = 9 + 3y. \\ \tt \implies x = \frac{9 + 3y}{5} - - - (3)$

Putting equation (3) in equation (1), we get

$\tt \implies3( \frac{ 9 + 3y}{5}) + 2y = 13.$

$\tt \implies \frac{27 + 9y + 10y}{5} = 13.$

$\tt \implies27 + 19y = 65.$

$\tt \implies19y = 38.$

$\tt \implies y = 2.$

Now, Putting the value of y in equation 3, we get.

$\tt \implies x = \frac{9 + 3(2)}{5}$

$\tt \implies x = \frac{15}{5}$

$\tt \implies x = 3.$

Therefore, the value of x =3 and y= 2.

Answer 2

$\huge{\bold{ AnSwEr }}$

x = 3 and y = 2

$\huge{ \underline{ \underline{ \red{ \sf{ SoLuTiOn :-}}}}}$

• By substitution method =>

3x + 2y = 13--------------(1)

5x - 3y = 9----------------(2)

from equ. (1) =>

3x + 2y = 13

3x = 13 - 2y

$x = \frac{13 - 2y}{3}$

$putting \: x = \frac{13 - 2y}{3} \: in \: equ. \: (2) \\ \\ 5x - 3y = 9 \\ \\ 5( \frac{13 - 2y}{3} ) - 3y = 9 \\ \\ \frac{65 - 10y}{3} - 3y = 9 \\ \\ 65 - 10y - 9y = 27 \\ \\ 65 - 19y = 27 \\ \\ - 19y = 27 - 65 \\ \\ - 19y = - 38 \\ \\ y = \frac{38}{19} \\ \\ y = 2$

Then the value of x =>

$x = \frac{13 - 2y}{3} \\ \\ x = \frac{13 - 2 \times 2}{3} \\ \\ x = \frac{9}{3} \\ \\ x = 3$

• By elimination method =>

3x + 2y = 13--------------(1)

5x - 3y = 9 ---------------(2)

Multiply by 5 in equ. (1) and 3 in equ. (2) =>

15x + 10y = 65 -----------(3)

15x - 9y = 27 -------------(4)

Subtracting equ. (3) and (4) =>

15x + 10y - ( 15x - 9y ) = 65 - 27

15x + 10y - 15x + 9y = 38

19y = 38

y = 38/19

y = 2

putting y = 2 in equ. (1) =>

3x + 2×2 = 13

3x = 13 - 4

3x = 9

x = 3

Here ,

${\purple{\boxed{\large{\bold{x = 3 \: and \: y = 2}}}}}$