Question

solve the following linear equations in two vairables by substitution method 2a+3b=13&5a-4y=-2​

Answer 1

Question:

solve the following linear equations in two vairables by substitution method $2a+3b=13 , 5a-4b= -2$

Answer:

A=2

B=3

Explanation:

The first equation is $2a+3b=13$ ---1

The Second equation is $5a -4b = -2$  

Let us solve on the first equation

⇒$2a+3b=13$

⇒$2a=13-3b$

⇒$a=13-3b/2$ -- 3

⇒$a=13-3b/2$ This is the third equation

Hence we have got the a we will substitute the 3rd equation in the 2nd

equation.

⇒$5a -4b = -2,a=13-3b/2$

⇒$5(13-3b/2)-4b=-2$

⇒$65-15b/2 (- 4b) = 2$

By taking L.C.M of -4b

⇒$65-15b-8b/2=-2$

By taking 2 from Left side to Right side it becomes multiplication

⇒$65-15b-8b=-2*2$

⇒$65-23b=(-4)$

⇒$-23b=-4-65$

⇒$-23b=-69$

- Minus sign will cancel due to both sides has negative power

⇒$23b=69$

⇒$b=69/23$

⇒$b=3$

If you have doubt in Substitution Process you can refer to Picture.

Hence B value is 3

Hence we can find the value of a by substituting the value of b that is 3 in equation 1

⇒$2a+3b=13,B=3$

⇒$2a+3(3)=13$

⇒$2a+9=13$

⇒$2a=13-9$

⇒$2a=4$

⇒$a=4/2$

⇒$a=2$

Hence the value of A is 2

Hence the value of B is 3