# Thread: System of equations

1. ## System of equations

Consider the system of equations:
ax + by = c, where a, b, c are in arithmetic sequence
dx + ey = f, where d, e, f are in arithmetic sequence

Given that ae ≠ bd, show that the system has x = -1 and y = 2 as its unique solution.

I've simplified the given system of equations and obtained an expression for 'x' and 'y' in terms of e, a, b, c , f and d. Yet, I'm unable to show that x=-1 and y=2 form a unique solution set. kindly help me with this.

MAX

2. ax + by = c, where a, b, c are in arithmetic sequence .......(1)
dx + ey = f, where d, e, f are in arithmetic sequence .......(2)
a, b, c are in arithmetic sequence $\displaystyle \therefore b-a=c-b$
or $\displaystyle b=\frac{(c+a)}{2} \quad similarly \quad e=\frac{(f+d)}{2}$
now on substituting values of b & e in (1) and (2), we get
$\displaystyle 2ax+(c+a)y=2c \quad ....(3)$
$\displaystyle 2dx+(f+d)y=2f \quad ....(4)$
solve (3) and (4) to get x=-1 and y=2.
for nature of solutions:- for unique solution

ratio of coefficient of x isn't equal to \ratio of coefficient of x

$\displaystyle \frac{a}{d} \neq \frac{b}{e}$
$\displaystyle or\quad ae \neq bd \quad(which\ is\ already\ given\ in\ question)$
therefore system of linear eq have unique solution x=-1 and y=2.

3. Originally Posted by MAX09
Consider the system of equations:
ax + by = c, where a, b, c are in arithmetic sequence
dx + ey = f, where d, e, f are in arithmetic sequence

Given that ae ≠ bd, show that the system has x = -1 and y = 2 as its unique solution.

I've simplified the given system of equations and obtained an expression for 'x' and 'y' in terms of e, a, b, c , f and d. Yet, I'm unable to show that x=-1 and y=2 form a unique solution set. kindly help me with this.

Since the problem is just to show that x= -1 and y= 2 are the solution (as long as $\displaystyle ae\ne bd$, the determinant of the coefficient matrix, ae- be, is not 0 so there is a unique solution), just put those into the equations and see what happens: a(-1)+ b(2)= c and d(-1)+ e(2)= f. The first is -a+ 2b= c . If b- a= r, then c= r+ a so -a+ 2b= -a+ 2(r+ a)= -a+ 2r+ 2a= a+ 2r, exactly what we should get since "a, b, c are in arithmetic sequence".