For what values of a and b will the system have infinitely many solutions 2x 3y 7?

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Given pair of linear equations is

2x + 3y = 7

2px + py = 28 – qy

or 2px + (p + q)y – 28 = 0

On comparing with ax + by + c = 0,

We get,

Here, a1 = 2, b1 = 3, c1 = – 7;

And a2 = 2p, b2 = (p + q), c2 = – 28;

a1/a2 = 2/2p

b1/b2 = 3/ (p+q)

c1/c2 = ¼

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a1/a2 = b1/b2 = c1/c2

1/p = 3/(p+q) = ¼

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

3/(p+q) = ¼

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

a1/a2 = 2/(a-b) ,               b1/b2 = 3/(a+b) ,               c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b

a – 9b = -4  ……………………………….(i)

2/(a-b) = 3/(a+b)

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

Given:

Equation 1: 2x + 3y = 7

Equation 2: 2ax + (a + b)y = 28

Both the equations are in the form of :

a1x + b1y = c1 & a2x + b2y = c2 where

a1 & a2 are the coefficients of x

b1 & b2 are the coefficients of y

c1 & c2 are the constants

For the system of linear equations to have infinitely many solutions we must have

According to the problem:

a1 = 2

a2 = 2a

b1 = 3

b2 = (a + b)

c1 = 7

c2 = 28

Putting the above values in equation (i) we get:

To obtain the value of a & b we need to solve the above equality. First we solve the extreme left and extreme right of the equality to obtain the value of a.

⇒ 

⇒ 2a x 7 = 2 x 28

⇒ 14a = 56

⇒ a = 4

After obtaining the value of a we again solve the extreme left and middle portion of the equality (ii)

⇒ 2 x (4 + b) = 3 x 2 x 4

⇒ b + 4 = 12

⇒ b = 8

The value of a & b for which the system of equations has infinitely many solution is a = 4 & b = 8

The given system of equations may be written as

The given system of equations will have infinite number of solutions, if

Hence, the given system of equations will have infinitely many solutions, if a=-5 and b=-1

The given system of equations can be written as
2x - 3y = 7
⇒2x - 3y - 7 = 0                        ….(i)
and (a + b)x - (a + b – 3)y = 4a + b
⇒(a + b)x - (a + b – 3)y - 4a + b = 0               ….(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
Here,` a_1 = 2, b_1= -3, c_1= -7 and a_2 = (a + b), b_2= -(a + b - 3), c_2= -(4a + b)`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`2/(a+b) = (−3)/(−(a+b−3)) = (−7)/(−(4a+b))`
`⇒ 2/(a+b) = 3/((a+b−3)) = 7/((4a+b))`
`⇒ 2/(a+b) = 7/((4a+b)) and 3/((a+b−3)) = 7/((4a+b))`
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b - 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b - 21
⇒ 4a = 17 and 5b = 11
∴ a = 5b                              …….(iii)
and 5a = 4b – 21               ……(iv)
On substituting a = 5b in (iv), we get:

25b = 4b – 21
⇒21b = -21
⇒b = -1
On substituting b = -1 in (iii), we get:
a = 5(-1) = -5
∴a = -5 and b = -1.

For what values of a and b will the system have infinitely many solutions?

Which is the solution of the equation 2x 3y 7?

What is the formula for infinitely many solutions?

For what values of A and B does the following pair of linear equations have infinite solutions?