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