Question

Identify which equations have one solution, infinitely many solutions, or no solution.

Answer 1

Let's label them 1-6, for convenience.

1. 1 solution (a positive # of ys equaling a positive constant)

2. Infinitely many solutions (You get 0=0 if you completely simplify)

3. No solution (take out the 3z's; 2.5 doesn't equal 3.2)

4. Infinitely many solutions (take out the 3/4 x; 1.1+2 = 3.1)

5. No solution (take out the 4.5r; 0 doesn't equal 3.2)

6. 1 solution (x = 3 1/2)

Answer 2

1)
`(1)/(2)y+(32)/(10)y=20` One solution

2)
`(15)/(2)+2z-(1)/(4)=4z+(29)/(4)-2z` Infinite Number of Solutions

3)
`3z+2.5=3.2+3z` No solution.

4)
`1.1+(3)/(4)x+2=3.1+(3)/(4)x` Infinitely Many Solutions

5)
`4.5r=3.2+4.5r` No solution

6)
`2x+4=3x+(1)/(2)` One solution

Explanation:

Equations may have exactly one solution, uncountable solutions or even no possible solution when the solution is a contradiction and this solution is never true.

1)
`(1)/(2)y+(32)/(10)y=20` One solution Let's prove it by solving it:

`(1)/(2)y+(32)/(10)y=20\Rightarrow (37)/(10)y=20\Rightarrow 10*(37)/(10)y=20*10\\37y=200\Rightarrow (37)/(37)y=(200)/(37)\Rightarrow y=(200)/(37)\Rightarrow S=\left \{ (200)/(37) \right \}`

2)
`(15)/(2)+2z-(1)/(4)=4z+(29)/(4)-2z` Infinite Number of Solutions because infinitely many solutions satisfies for z.

`(15)/(2)+2z-(1)/(4)=4z+(29)/(4)-2z\\(29)/(4)+2z=(29)/(4)+2z`

3)
`3z+2.5=3.2+3z` No solution. There's no way to add 2.5 to 3z and have the same amount as adding 3.2 to 3z. This is contradiction. This is a false equality.

4)
`1.1+(3)/(4)x+2=3.1+(3)/(4)x` Infinitely many solutions. This equation has infinitely many solutions since the left side is equal to the right side, any value plugged in x may result in many solutions.

5)
`4.5r=3.2+4.5r` No solution Similarly, again. There's no way of adding 3.2 to 4.5r being equal to 4.5r. Another contradiction. This is a false equality.

6)
`2x+4=3x+(1)/(2)` One solution

`2x+4=3x+(1)/(2)\\2x+4-2x=3x+(1)/(2)-2x\\4=x+(1)/(2)\\4-(1)/(2)=x\Rightarrow x=(7)/(2)\Rightarrow S=\left \{ (7)/(2) \right \}`

Since we can see on the left side different expressions than on the right side. All that is left is doing the test, by solving it.