Question

What values of a and b make this system of equations have no solution ?

Answer 1

Step 1. Two equations have no solutions if the lines that they represent are parallel lines:

And to be parallel lines they must have the same slope and different y-intercepts.

Step 2. The two equations that we have are:

`\begin{gathered} 3x+4y=-4 \\ y=ax+b \end{gathered}`

The second equation

y=ax+b

is already in the slope-intercept form where a is the slope and b is the y-intercept of the line.

The first equation

3x+4y=-4

is not in the slope-intercept form. Therefore, we solve for y:

`\begin{gathered} 3x+4y=-4 \\ \downarrow \\ 4y=-3x-4 \\ \downarrow \\ y=(-3)/(4)x-1 \end{gathered}`

Step 3. The pair of equations now is:

`\begin{gathered} y=(-3)/(4)x-1 \\ y=ax+b \end{gathered}`

For the lines to be parallel, the slope of the two lines must be the same.

The slope of the first line is -3/4, that is the value that a should have:

`a=-(3)/(4)`

And the value of b cannot be -1 because then the equation would be the same and instead of no solutions there would be infinite solutions. Therefore a possible value can be b=1

`b=1`

These