Question

For what value of a and b would the following pairs of linear equations have infinite solutions.
8x=10-6y,4x+(a-b)y=(a+b-2)

Answer 1

$\fbox{\texttt{\green{Answer:}}}$

• Value of a = 5
• Value of b = 2

$\fbox{\texttt{\pink{Given:}}}$

The pair of linear equations have infinite number of solutions.

$\fbox{\texttt{\orange{To\:find:}}}$

Value of a & b.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

We are given two linear equations,

• 8x + 6y - 10 = 0 ------(1)
• 4x + (a + b)y -(a + b -2) = 0 ------(2)

For infinite number of solutions, we have,

$\rm{\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}}$

Now we have,

$\rm{\dfrac{a_1}{a_2}=\dfrac{8}{4}=2}$

Again,

$\rm{\dfrac{b_1}{b_2}=\dfrac{6}{(a-b)}}$

Again,

$\rm{\dfrac{c_1}{c_2}=\dfrac{-10}{-(a+b-2)}}$

$\hrulefill$

Applying the formula, we get,

$\rm{2=\dfrac{6}{a-b}=\dfrac{10}{a+b-2}}$

Now,

⇝ 2 = 6/a - b

⇝ 2a - 2b = 6

⇝ a - b = 3 -------(3)

Again,

⇝ 2 = 10/a + b -2

⇝ 2a + 2b -4 = 10

⇝ a + b -2 = 5

⇝ a + b = 7 -------(4)

Adding (3) & (4), we get,

⇝ 2a = 10

⇝ a = 5

Substituting the value of a in (4), we get,

⇝ 5 + b = 7

⇝ b = 2

∴ Value of a = 5 & value of b = 2.

Answer 2

QuEsTiOn :-

For what value of a and b would the following pairs of linear equations have infinite solutions

8x=10-6y,4x+(a-b)y=(a+b-2).

$\huge{ \underline{ \underline{ \red{ \sf{ SoLuTiOn :-}}}}}$

8x = 10 - 6y

8x + 6y - 10 = 0------------(1)

4x + (a-b)y = a + b - 2

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

For Infinite Solution :-

$\frac{a_{1}}{a_{2} } = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ \\ Here, \: \\ a_{1} = 8, \: a_{2} = 4, \: b_{1} = 6, \: b_{2} = a + b ,\: \\ c_{1} = - 10 ,\: c_{2} = - (a + b - 2) \\ \\ Now, \: \\ \frac{a_{1}}{a_{2}} = \frac{8}{4} = 2 \\ \\ \frac{b_{1}}{b_{2}} = \frac{6}{a - b} \\ \\ \frac{c_{1}}{c_{2}} = \frac{ - 10}{ - (a + b - 2)} = \frac{10}{a + b - 2} \\ \\ then \: \implies\\ 2 = \frac{6}{a - b} = \frac{10}{a + b - 2} \\ \\ Comparing \: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \\ \\ 2 = \frac{6}{a - b} \\ \\ 2a - 2b = 6 \\ \\ a - b = 3 - (3) \\ \\ again \: \\ \frac{a_{1}}{a_{2}} = \frac{c_{1}}{c_{2}} \\ \\ 2 = \frac{10}{a + b - 2} \\ \\ 2a + 2b - 4 = 10 \\ \\ 2a + 2b = 14 \\ \\ a + b = 7 - (4) \\ \\ Subtracting \: equ. \: (3) \: and \: (4) \\ \\ a - b - (a + b) = 3 - 7 \\ \\ a - b - a - b = - 4 \\ \\ - 2b = - 4 \\ \\ 2b = 4\\ \\ b = 2 \\ \\ putting \: b = 2 \: in \: equ.(3) \\ \\ a - 2 = 3 \\ \\ a = 3 + 2 \\ \\ a = 5$

AnSwEr :-

${\blue{\boxed{\large{\bold{a = 5 \: and \: b = 2 }}}}}$