Question

2x-3y=7,(a+b+1)x+(a+2b+2)y=4(a+b)+1.​

Answer 1

Answer:

The solution is

$x=\frac{19a+26b+17}{5a+7b+7}$

$y=\frac{a+b-5}{5a+7b+7}$

Step-by-step explanation:

I have solved the system of linear equations by cross multiplication rule.

Given equations are

$2x-3y-7=0$

$(a+b+1)x+(a+2b+2)y-7-4(a+b)-1=0$

By cross multiplication rule,

$\frac{x}{12(a+b)+3+7(a+2b+2)}=\frac{y}{-7(a+b+1)+8(a+b)+2}=\frac{1}{2a+4b+4+3a+3b+3}$

$\frac{x}{12a+12b+3+7a+14b+14}=\frac{y}{-7a-7b-7+8a+8b+2}=\frac{1}{5a+7b+7}$

$\frac{x}{19a+26b+17}=\frac{y}{a+b-5}=\frac{1}{5a+7b+7}$

$\frac{x}{19a+26b+17}=\frac{1}{5a+7b+7}$

$x=\frac{19a+26b+17}{5a+7b+7}$

$\frac{y}{a+b-5}=\frac{1}{5a+7b+7}$

$y=\frac{a+b-5}{5a+7b+7}$

The solution is

$x=\frac{19a+26b+17}{5a+7b+7}$

$y=\frac{a+b-5}{5a+7b+7}$

Answer 2

Answer:

a = 21 &

b = -16

Step-by-step explanation:

2x-3y=7,(a+b+1)x+(a+2b+2)y=4(a+b)+1.​

we need to find values of a & b so that these equations have infinite solutions

2/(a+b+1)  =  -3/(a+2b +2)  = 7/(4(a+b) + 1)

Equating 1st & 2nd

=> 2a + 4b + 4 = -3a -3b - 3

=> 5a + 7b = - 7   eq 1

Equating 1st & 3rd

8a + 8b + 2  = 7a + 7b + 7

=> a + b  = 5     eq 2

eq 1  - 5Eq2

=> 2b = -32

=> b = -16

putting in eq 2

a = 21

a = 21 & b = -16 will give infinite solution

with these values

(a+b+1)x+(a+2b+2)y=4(a+b)+1  becomes

6x - 9y = 21

Diving by 3

2x - 3y = 7

Same as 1st Equation