# Math Help - 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.

Thanks in advance,
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 $\therefore b-a=c-b$
or $b=\frac{(c+a)}{2} \quad similarly \quad e=\frac{(f+d)}{2}$
now on substituting values of b & e in (1) and (2), we get
$2ax+(c+a)y=2c \quad ....(3)$
$2dx+(f+d)y=2f \quad ....(4)$
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

$\frac{a}{d} \neq \frac{b}{e}$
$or\quad ae \neq bd \quad(which\ is\ already\ given\ in\ question)$
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.

Thanks in advance,
MAX
To "show that .... satisfy equation..." it is not necessary to solve the equation. It is only necessary to put the purported solution into the equation and show that it is satisfied.

Since the problem is just to show that x= -1 and y= 2 are the solution (as long as $ae\ne bd$, 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".

Do the same thing with the second equation.