Question

If system of equation 2x+3y=7 and (a+b)x+(2a-6)y=21 has infinitely many solutions then find a and b

Answer 1

Hello !

2x + 3y = 7
2x + 3y  -7 = 0
a₁ = 2 , b₁ = 3 , c₁ = -7

(a+b)x+(2a-b)y = 21
(a+b)x+(2a-b)y - 21= 0

a₂ = (a+b)  , b₂ = (2a - b) , c₂ = -21

------------------------------------------
As the equations have infinite solutions:-
a₁/a₂ = b₁/b₂ = c₁/c₂


$\frac{2}{a+b} = \frac{3}{2a - b} = \frac{-7}{-21}$


[tex] \frac{2}{a + b} = \frac{7}{21} \\ 7a + 7b = 42 \\ \\ a + b = 6 [/tex]   ----> (1)

-------------------------------------------------------------
[tex] \frac{3}{2a - b} = \frac{7}{21 } \\ \\ 7(2a-b) = 63 \\ [/tex]

2a - b = 9           ----> (2)

Adding equations 1 and 2 :-

a + b = 6
2a - b = 9
--------------
3a = 15

a = 5

a + b = 6
b = 1

------------------------------------------------

a = 5 , b = 1

Answer 2

Step-by-step explanation:

Hello !

2x + 3y = 7

2x + 3y -7 = 0

a₁ = 2 , b₁ = 3 , c₁ = -7

(a+b)x+(2a-b)y = 21

(a+b)x+(2a-b)y - 21= 0

a₂ = (a+b) , b₂ = (2a - b) , c₂ = -21

------------------------------------------

As the equations have infinite solutions:-

a₁/a₂ = b₁/b₂ = c₁/c₂

$\frac{2}{a+b} = \frac{3}{2a - b} = \frac{-7}{-21}$

$\begin{lgathered}\frac{2}{a + b} = \frac{7}{21} \\ 7a + 7b = 42 \\ \\ a + b = 6\end{lgathered} </p><p>$

-------------------------------------------------------------

$\begin{lgathered}\frac{3}{2a - b} = \frac{7}{21 } \\ \\ 7(2a-b) = 63 \\\end{lgathered}.$

7(2a−b)=63

2a - b = 9 ----> (2)

Adding equations 1 and 2 :-

a + b = 6

2a - b = 9

--------------

3a = 15

a = 5

a + b = 6

b = 1

------------------------------------------------

a = 5 , b = 1