Question

If we have $a < b < c < d$. How do you find the solution of $\left| x-a \right|+\left| x-c \right|=\left| x-b \right|+\left| x-d \right|$? Is it verifiable that for $\left( a,b,c,d \right)=\left( 1,2,3,5 \right),x=\dfrac{7}{2}$ is a solution.

Answer 1

To solve the equation we will take different cases where x lies in the interval $x < a,a < x < b,b < x < c,c < x < d$ and $x>d$ . Hence using this cases we will open the modulus function and find the condition obtained. Now we will substitute values to check if the conditions obtained are possible.

Complete step-by-step answer:
Now we know that the modulus function $\left| x \right|$ is defined as $x,x>0$ and $-x,x < 0$
Now let us consider that x < a.
Then we have $x-a < 0,x-b < 0,x-c < 0$ and $x-d < 0$
Hence opening the modulus we get,
$\begin{aligned} & \Rightarrow -\left( x-a \right)-\left( x-c \right)=-\left( x-b \right)-\left( x-d \right) \\\ & \Rightarrow a-x+c-x=b-x+d-x \\\ & \Rightarrow a+c=b+d \\\ \end{aligned}$
Now since we have $a < b < c < d$ we cannot have $a+c=b+d$ hence this case will never be possible
Now consider the case where $a < x < b$ .
Then we have, $x-b < 0,x-c < 0,x-d < 0$ and $\left( x-a \right)>0$
Hence opening the modulus we have,
$\begin{aligned} & \Rightarrow \left( x-a \right)-\left( x-c \right)=-\left( x-b \right)-\left( x-d \right) \\\ & \Rightarrow x-a-x+c=b-x+d-x \\\ & \Rightarrow 2x=-c+d+b+a \\\ & \Rightarrow x=\dfrac{a+b+d-c}{2} \\\ \end{aligned}$
Now we can substitute the values and check that this cannot be true for any x between a and b.
Now consider the case where $b < x < c$ .
Then we have $x-a>0,x-b>0$ and $x-c < 0,x-d < 0$ .
Hence we get the equation as,
$\begin{aligned} & \Rightarrow \left( x-a \right)-\left( x-c \right)=\left( x-b \right)-\left( x-d \right) \\\ & \Rightarrow x-a-x+c=x-b-x+d \\\ & \Rightarrow c-a=d-b \\\ \end{aligned}$
$\Rightarrow b+c=a+d$
Now again this will not be possible as $a < b < c < d$
Now consider the case $c < x < d$
Then only $x-d < 0$ rest all are positive. Hence we get,
$\begin{aligned} & \Rightarrow x-a+x-c=x-b-\left( x-d \right) \\\ & \Rightarrow x-a+x-c=x-b+d-x \\\ & \Rightarrow 2x=a+c+d-b \\\ & \Rightarrow x=\dfrac{a+c+d-b}{2} \\\ \end{aligned}$
Now on substituting $\left( a,b,c,d \right)=\left( 1,2,3,5 \right)$ we get $x=\dfrac{7}{2}$ Hence this is the solution.
And if $x>d$ then all the terms will be positive and hence we have,
$\begin{aligned} & \Rightarrow x-a+x-c=x-b+x-d \\\ & \Rightarrow -a-c=-b-d \\\ & \Rightarrow a+c=b+d \\\ \end{aligned}$
Now since we have $a < b < c < d$ we cannot have $a+c=b+d$ hence this case will never be possible.
Hence the solution of the given equation is $\left( a,b,c,d \right)=\left( 1,2,3,5 \right)$ and $x=\dfrac{7}{2}$

Note: Now note that we can also solve the problem by drawing the graph of two function $\left| x-a \right|+\left| x-c \right|$ and the function $\left| x-b \right|+\left| x-d \right|$ . To draw the function we will have to open the modulus by considering different cases. Now we will plot the two functions and find the intersection. Hence we can find the solution of the given equation.