Given that the equations in $x$ :- $3[x - 2(x + \frac{a}{3})] = 2x$ and $\frac{3x + a}{3} - \frac{1 + 4x}{6} = 0$ has a common solution , find it .

Question

Given that the equations in $x$ :- $3[x - 2(x + \frac{a}{3})] = 2x$ and $\frac{3x + a}{3} - \frac{1 + 4x}{6} = 0$ has a common solution , find it .

What I Tried :- We have :-

$\frac{3x + a}{3} - \frac{1 + 4x}{6} = 0$

=> $\frac{6x + 2a - 1 - 4x}{6} = 0$

=> $2x + 2a - 1 = 0$

=> $x = \frac{1 - 2a}{2}$ .

Substitute $x$ in the $1$st equation to get :-

$3[x - 2(x + \frac{a}{3})] = 1 - 2a$

=> $3[-x - \frac{2a}{3}] = 1 - 2a$

=> $-3x - 2a = 1 - 2a$

=> $x = -\frac{1}{3}.$

I got the value of $x$ , and I suppose I will find $a$ too , but how to find the common solution?

Can anyone help me?

Answer 1

You already computed $x=-\frac 13$ and $x=\frac{1-2a}2$. From these:$$-\frac13=\frac12-a$$

Answer 2

After getting $x=\frac {1-2a}2$ from the second equation you get $3[(\frac {1-2a}2-2)(\frac {1-2a}2+\frac a 3)]=2\frac {1-2a}2$ from the first equation. Solve this for $a$.