Question

Find the values of a and b for which the system of equations has infinitely many

solutions

3x(a + 1)y = 2b - 1 and 5x + (1 - 2a)y = 3b​

Answer 1

Step-by-step explanation:

Correct Equations:-

3x - (a + 1)y = 2b - 1 and 5x + (1 - 2a)y = 3b

Given:-

• Two equations:-

1. 3x - (a + 1)y = 2b - 1
2. 5x + (1 - 2a)y = 3b

To Find:-

• The values of a and b

Concept used:-

Suppose we have two equations in two variables

$\bull \sf \: a_{1}x + b_{1}y = c_{1} \\ \bull \sf \: a_{2}x + b_{2}y = c_{2}$

Given two equations have infinitely many solutions only if:-

$\boxed{\sf \leadsto \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}}$

Solution:-

$\sf 3x - (a + 1)y = 2b - 1 \: \: \: \: \: \: \: \: \big[a_{1} = 3 \: , \: b_{1} = - (a + 1) \: , \: c_{1} = 2b - 1 \big]$

$\sf 5x + (1 - 2a)y = 3b \: \: \: \: \: \: \: \: \big[a_{2} = 5 \: , \: b_{2} = 1 - 2a\: , \: c_{1} = 3b \big]$

Since, the two equations has infinitely many solutions.

$\sf \implies\dfrac{3}{5} = \dfrac{ - (a + 1)}{1 - 2a} = \dfrac{2b - 1}{3b}$

$\:$

$\sf \dfrac{3}{5} = \dfrac{ - (a + 1)}{1 - 2a} \: \:(and) \: \: \dfrac{3}{5} = \dfrac{2b - 1}{3b}$

$\:$

$\sf \implies\dfrac{3}{5} = \dfrac{ - (a + 1)}{1 - 2a}$

$\sf \implies3(1 - 2a)= -5 (a + 1)$

$\sf \implies3 - 6a= -5a - 5$

$\sf \implies6a - 5a= 5 + 3$

$\sf \implies \underline{ \: \: \underline{ \: a = 8 \: } \: \: }$

Now:-

$\sf\implies \dfrac{3}{5} = \dfrac{2b - 1}{3b}$

$\sf\implies 3(3b)= 5(2b - 1)$

$\sf\implies 9b= 10b - 5$

$\sf\implies 10b - 9b = 5$

$\sf \implies \underline{ \: \: \underline{ \: b = 5\: } \: \: }$

$\large\underline{\boxed{\therefore \textsf{\textbf{ a = 8 \: \: , \: \: b = 5}}}}$

Answer 2

$\huge{\boxed{\rm{Correct \: question}}}$

Find the values of a and b for which the system of equations has infinitely many solutions →

• 3x - (a + 1)y = 2b - 1 and 5x + (1 - 2a)y = 3b

$\huge{\boxed{\rm{Answer}}}$

$\large{\boxed{\boxed{\sf{Given \: that}}}}$

Both equations are given below ➼

• 3x - (a + 1)y = 2b - 1

• 5x + (1 - 2a)y = 3b

$\large{\boxed{\boxed{\sf{To \: find}}}}$

• The value of a and b.

$\large{\boxed{\boxed{\sf{Solution}}}}$

• Value of a = 8

• Value of b = 5

$\large{\boxed{\boxed{\sf{Using \: formula \: or \: concept}}}}$

Let we have 2 equations in 2 variables

➸ ᵃ1ˣ = ᵇ1ʸ = ᶜ1

➸ ᵃ2ˣ = ᵇ2ʸ = ᶜ2

Now,

ᵃ1 / ᵃ2 = ᵇ1 / ᵇ2 = ᶜ1 / ᶜ2

$\large{\boxed{\boxed{\sf{What \: does \: the \: question \: says}}}}$

$\large{\boxed{\boxed{\sf{Let's \: understand \: the \: concept \: now}}}}$

• This question have equation 3x - (a + 1)y = 2b - 1 and 5x + (1 - 2a)y = 3b in these equations we have to find the value of a and b.

$\large{\boxed{\boxed{\sf{How \: to \: do \: this \: question}}}}$

$\large{\boxed{\boxed{\sf{Let's \: see \: the \: procedure \: now}}}}$

• To solve this question we have to use formula afterthat putting the values we get the value of a that is 8 afterthat substituting the value we get value of b as 5.

$\large{\boxed{\boxed{\sf{Full \: solution}}}}$

☆ 3x - (a + 1)y = 2b - 1

Where the values are

• ᵃ1 = 3

• ᵇ1 = -a+1

• ᶜ1 = 2b - 1

☆ 5x + (1-2a)y = 3b

Where the values are

• ᵃ1 = 5

• ᵇ2 = 1

• ᶜ2 = 3b

Since, both of the solutions have infinite solutions .

• 3/5 = (-a+1) / (1-2a) and 3/5 = 2b - 1 / 3b.

• 3/5 = (-a+1) / (1-2a) and 3/5 = 2b - 1 / 3b.

• 3/5 = (-a+1) / (1-2a).

Cross multiplying the digits we get

• 3/5(1-2a) = -5(a+1)

• 3-6a = -5a - 5

Placing like terms we get

• 6a - 5a = 5 + 3

• 1a = 8

• a = 8

Hence, value of a = 8

Now,

• 3/5 = 2b-1 / 3b

Cross multiplying the digits we get

• 3(3b) = 5(2b-1)

• 9b = 10b - 5

• 10b - 9b = 5

• 1b = 5

• b = 5

Hence, value of b = 5