Question

Consider the equation

2(5x - 4) = ax + b

Find a value for a and b so that the equation has no solution.

I’ll give you 10 points please I really need help!

Answer 1

To find values for a and b so that the equation has no solution, we need to make sure that the equation is in the form of a quadratic equation and that the discriminant (b^2 - 4ac) is negative. In this case, there is no combination of values for a and b that will make the equation have no solution.

Step-by-step explanation:

To find values for a and b so that the equation has no solution, we need to make sure that the equation is in the form of a quadratic equation and that the discriminant (b^2 - 4ac) is negative. In the given equation, 2(5x - 4) = ax + b, we can rewrite it as 10x - 8 = ax + b. Comparing it with the quadratic equation form (ax^2 + bx + c = 0), we can see that a = 0, b = 10, and c = -8. Now, we can calculate the discriminant: b^2 - 4ac = 10^2 - 4(0)(-8) = 100. Since the discriminant is positive, there will be solutions for this equation. Therefore, there is no combination of values for a and b that will make the equation have no solution.

Answer 2

A.
`a=10\\ \\b\\eq -8`

B.
`a=10\\ \\b= -8`

Step-by-step explanation:

Consider the equation

`2(5x-4) = ax + b`

A. This equation has no solutions when the coefficients at x are the same and the free coefficients are not the same.

First, use distributive property:

`2(5x-4)=2\cdot 5x-2\cdot 4=10x-8`

So, the equation is

`10x-8=ax+b`

This equation has no solutions when

`a=10\\ \\b\\eq -8`

B. The equation has infinitely many solutions when the coefficients at x are the same and the free coefficients are the same too.

So, the equation

`10x-8=ax+b`

has infinitely many solutions when

`a=10\\ \\b= -8`

In other cases, the equation has a unique solution