Respuesta :
we have
[tex]\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x[/tex]
Combine like terms in both sides
[tex](\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x[/tex]
we know that
[tex](\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x[/tex]
substitute in the expression above
[tex]\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x[/tex]-----> equation A
Multiply equation A by [tex]5*3*2=30[/tex] both sides
[tex]30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)[/tex]
[tex]48x+20=15-6x[/tex] ---------> equation B
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]48x+6x=15-20[/tex]
[tex]54x=-5[/tex] ---------> equation C
Solve for x
[tex]x=-\frac{5}{54} =-0.09[/tex]
We are going to proceed to verify each case to determine the solution.
Case a) [tex]\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x[/tex]
the case a) is equal to the equation A
so
the case a) have the same solution that the given equation
Case b) [tex]18x+20+30x=15-6x[/tex]
Combine like terms in left side
[tex](18x+30x)+20=15-6x[/tex]
[tex](48x)+20=15-6x[/tex]
the case b) is equal to the equation B
so
the case b) have the same solution that the given equation
Case c) [tex]18x+20+x=15-6x[/tex]
Combine like terms in left side
[tex](18x+x)+20=15-6x[/tex]
[tex](19x)+20=15-6x[/tex]
[tex]19x+6x=15-20\\25x=-5\\x=-0.20[/tex]
[tex]-0.20\neq -0.09[/tex]
therefore
the case c) not have the same solution that the given equation
Case d) [tex]24x+30x=-5[/tex]
Combine like terms in left side
[tex]54x=-5[/tex]
the case d) is equal to the equation C
so
the case d) have the same solution that the given equation
Case e) [tex]12x+30x=-5[/tex]
Combine like terms in left side
[tex]42x=-5[/tex]
[tex]x=-5/42=-0.12[/tex]
[tex]-0.12\neq -0.09[/tex]
therefore
the case e) not have the same solution that the given equation
therefore
the answer is
case a) [tex]\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x[/tex]
case b) [tex]18x+20+30x=15-6x[/tex]
case d) [tex]24x+30x=-5[/tex]
Answer:
Option (a) , (b) and ( d) are equivalent to given expression [tex]\frac{3}{5}x+\frac{2}{3}+x= \frac{1}{2}- \frac{1}{5}x[/tex]
Step-by-step explanation:
Given equation : [tex]\frac{3}{5}x+\frac{2}{3}+x= \frac{1}{2}- \frac{1}{5}x[/tex]
We have to use properties to rewrite the given equation and check which are correct from thee given options,
Consider the given equation,
[tex]\frac{3}{5}x+\frac{2}{3}+x= \frac{1}{2}- \frac{1}{5}x[/tex]
Applying commutative property of addition , [tex]a+b=b+ a[/tex]
Equation becomes,
[tex]\frac{3}{5}x+x+\frac{2}{3}= \frac{1}{2}- \frac{1}{5}x[/tex]
Now adding x terms on right side , we get,
[tex]\frac{3+5}{5}x+\frac{2}{3}= \frac{1}{2}- \frac{1}{5}x[/tex]
[tex]\Rightarrow \frac{8}{5}x+\frac{2}{3}= \frac{1}{2}- \frac{1}{5}x[/tex]
Thus, obtained option (a).
Again consider given equation ,
[tex]\frac{3}{5}x+\frac{2}{3}+x= \frac{1}{2}- \frac{1}{5}x[/tex]
Taking LCM both sides, we get,
[tex]\frac{9x+10+15x}{15}= \frac{5-2x}{10}[/tex]
Solving , we get,
[tex]\frac{9x+10+15x}{3}= \frac{5-2x}{2}[/tex]
Cross multiply, we get,
[tex]2\times (9x+10+15x)=3\times(5-2x)[/tex]
[tex]18x+20+30x=15-6x[/tex]
Thus, obtained option (b).
Taking like terms together,
[tex]18x+6x+30x=15-20[/tex]
[tex]\Rightarrow 24x+30x=-5[/tex]
Thus, obtained option (d).
Thus, Option (a) , (b) and ( d) are equivalent to given expression [tex]\frac{3}{5}x+\frac{2}{3}+x= \frac{1}{2}- \frac{1}{5}x[/tex]