What value of b will cause the system to have an infinite number of solutions? y = 6x – b –3x + 1/2y = –3

Verified Answer :

we have

\(y=6x-b\) -----> equation A

\(-3x+\frac{1}{2}y=-3\) -----> equation B

we know that

If the system has infinite number of solutions then, the equation A is equal to the equation B

so

isolate variable y in the equation B

\(-3x+\frac{1}{2}y=-3\)

Multiply by \(2\) both sides

\(-6x+y=-6\)  

\(y=6x-6\) -------> new equation B

To find the value of b, equate equation A and equation B

\(6x-b=6x-6\)

\(b=6\)

therefore

the answer is

\(b=6\)

The value of b should be 6 for the system of equations to give infinite solution.

What is System of equation?

Inconsistent System

A system of equations to have no real solution, the lines of the equations must be parallel to each other.

Consistent System

1. Dependent Consistent System

A system of the equation to be Dependent Consistent System the system must have multiple solutions for which the lines of the equation must be coinciding.

2. Independent Consistent System

A system of the equation to be Independent Consistent System the system must have one unique solution for which the lines of the equation must intersect at a particular.

Given to us

  1. \(y = 6x-b\)
  2. \(-3x+\dfrac{1}{2}y = -3\)

As we know for a system of equations to give an infinite number of solutions, the line of the equations must coincide, therefore, the equations must be equal or in ratio.

Equation 1

As the value of y is already mentioned in equation 1,

\(y = 6x-b\)

Equation 2

Substitute the value of y in equation 2,

\(-3x+\dfrac{1}{2}(6x-b) = -3\\\\-3x +(6x\dfrac{1}{2})-(b\dfrac{1}{2})=-3\\\\-3x+3x-(b\dfrac{1}{2})=-3\\\\(b\dfrac{1}{2})=3\\\\b=6\)

Hence, the value of b should be 6 for the system of equations to give infinite solution.

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