Question

2(a -3) + 3(b-5)=0...........equation1 5(a-1) + 4(b-4)...............equation2​

Answer 1

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Answer 2

The values of a and b are -3 and 9 respectively.

Step-by-step explanation:

We are given two linear equations,

$2(a-3)+3(b-5)=0$         --- (1)

$5(a-1)+4(b-4)=0$        ---  (2)

and we have to find the values of a and b.

• Calculation for 'a' and 'b'

First of all we do multiplication for opening the brackets,

from equation (1),

$2*a+2*(-3)+3*b+3*(-5)=0$

$2a-6+3b-15=0$

$2a+3b-21=0$      

$2a+3b=21$       ---(3)

from equation (2),

$5*a+5*(-1)+4*b+4*(-4)=0$

$5a-5+4b-16=0$

$5a+4b-21=0$

$5a+4b=21$        ----(4)

hence the R.H.S of equations (3) and (4) are same so we can equate L.H.S also in order to get the value of 'a'

$2a+3b=5a+4b$

taking similar terms one side.

$-b=3a$

$a=\frac{-b}{3}$    ----(5)

now substitute this 'a' in equation (3),

$2a+3b=21$    

$2*(\frac{-b}{3})+3b=21$

$-2b+3*3b=21*3$

$-2b+9b=63$

$7b=63$

$b=\frac{63}{7}$

$b=9$

we get the value of b as $b=9$. by putting this value of b in equation (5) we get ,

 $a=\frac{-b}{3}$    

    $=\frac{-9}{3}$

$a=-3$

The values of a and b are -3 and 9 respectively.