Math, asked by Anonymous, 10 months ago

solve it fast urgent

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Answered by bhavyagugnani

Answer:)

2x+3y-7=0-----(1)

(a-b)x+(a+b)y-(3a+b-2)=0-----(2)

Here, $a_{1}$ =2; $b_{1\\}$ =3; $c_{1}$ =-7

$a_{2}$ = (a-b); $b_{2}=(a+b)$ ; $c_{2}=-(3a+b-2)$

$\frac{a_{1}}{a_{2}}=\frac{2}{a-b}; \frac{b_{1}}{b_{2}} =\frac{3}{a+b} ; \frac{c_{1}}{c_{2}} =\frac{7}{3a+b-2}$

For the equations to have infinitely many solutions, we have:

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$Comparing \frac{a_{1}}{a_{2}} with \frac{b_{1}}{b_{2}}$

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

2(a+b)=3(a-b)----------------------(cross-multiplying)

2a+2b=3a-3b

2a-3a+2b+3b=0

-a+5b=0-------------------(3)

Comparing $\frac{b_{1}}{b_{2}}$ with $\frac{c_{1}}{c_{2}}$

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

3(3a+b-2)=7(a+b)----------------------(cross-multiplying)

9a+3b-6=7a+7b

9a-7a+3b-7b=6

2a-4b=6-------------(4)

Subtracting (1) from (2), we obtain:

4b = 4

b = 1

Substituting the value of b in equation (2), we obtain:

a - 5 x 1 = 0

a = 5

Thus, the values of a and b are 5 and 1 respectively.

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Answered by anmol7337

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2x+3y-7=0-----(1)

(a-b)x+(a+b)y-(3a+b-2)=0-----(2)

Here, $$a_{1}$$ =2; $$\begin{lgathered}b_{1\\}\end{lgathered}$$ =3; $$c_{1}$$ =-7

$$a_{2}$$ = (a-b); $$b_{2}=(a+b)$$ ; $$c_{2}=-(3a+b-2)$$

$$\frac{a_{1}}{a_{2}}=\frac{2}{a-b}; \frac{b_{1}}{b_{2}} =\frac{3}{a+b} ; \frac{c_{1}}{c_{2}} =\frac{7}{3a+b-2}$$

For the equations to have infinitely many solutions, we have:

$$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

$$Comparing \frac{a_{1}}{a_{2}} with \frac{b_{1}}{b_{2}}$$

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