Answer:
1) [tex]\frac{1}{2}y+\frac{32}{10}y=20[/tex] One solution
2)[tex]\frac{15}{2}+2z-\frac{1}{4}=4z+\frac{29}{4}-2z[/tex] Infinite Number of Solutions
3) [tex]3z+2.5=3.2+3z[/tex] No solution.
4) [tex]1.1+\frac{3}{4}x+2=3.1+\frac{3}{4}x[/tex] Infinitely Many Solutions
5) [tex]4.5r=3.2+4.5r[/tex] No solution
6) [tex]2x+4=3x+\frac{1}{2}[/tex] One solution
Step-by-step 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) [tex]\frac{1}{2}y+\frac{32}{10}y=20[/tex] One solution Let's prove it by solving it:
[tex]\frac{1}{2}y+\frac{32}{10}y=20\Rightarrow \frac{37}{10}y=20\Rightarrow 10*\frac{37}{10}y=20*10\\37y=200\Rightarrow \frac{37}{37}y=\frac{200}{37}\Rightarrow y=\frac{200}{37}\Rightarrow S=\left \{ \frac{200}{37} \right \}[/tex]
2)[tex]\frac{15}{2}+2z-\frac{1}{4}=4z+\frac{29}{4}-2z[/tex] Infinite Number of Solutions because infinitely many solutions satisfies for z.
[tex]\frac{15}{2}+2z-\frac{1}{4}=4z+\frac{29}{4}-2z\\\frac{29}{4}+2z=\frac{29}{4}+2z[/tex]
3) [tex]3z+2.5=3.2+3z[/tex] 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) [tex]1.1+\frac{3}{4}x+2=3.1+\frac{3}{4}x[/tex] 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) [tex]4.5r=3.2+4.5r[/tex] 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) [tex]2x+4=3x+\frac{1}{2}[/tex] One solution
[tex]2x+4=3x+\frac{1}{2}\\2x+4-2x=3x+\frac{1}{2}-2x\\4=x+\frac{1}{2}\\4-\frac{1}{2}=x\Rightarrow x=\frac{7}{2}\Rightarrow S=\left \{ \frac{7}{2} \right \}[/tex]
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.
