[Soln] 2014 Singapore MO (Junior Rd 1) Problem 31

2014 SMO (Junior-Rd1) 31. If ax + by = 7, ax^2 + by^2 = 49, ax^3 + by^3 = 133, ax^4 + by^4 = 406, find the value of 2014(x+y-xy) - 100(a+b).

Let’s label the equations so that it’ll be easier to refer to them:

\begin{cases} ax + by = 7 &- (1) \\  ax^2 + by^2 = 49 &- (2) \\  ax^3 + by^3 = 133 &-(3) \\  ax^4 + by^4 = 406 &-(4) \end{cases}

A good way to start making sense of simultaneous equations is to eliminate the variables, one at a time. The easiest way to do so is to multiply one equation by a suitable constant so that one of its terms matches one of the terms in another equation. (This is exactly what we do when we do Gaussian elimination for simultaneous linear equations.) For this question, we can multiply (1) by x so that it will have a ax^2 term to match (2). Performing (2) - x \cdot (1):

\begin{aligned} (ax^2 + by^2) - (ax^2 + bxy) &= 49 - 7x, \\  by(y-x) &= 49-7x. -(5) \end{aligned}

On its own it doesn’t seem like much, but we can exploit the symmetry we see in the given equations to get 2 similar equations by performing (3) - x \cdot (2) and (4) - x \cdot (3), resulting in 3 equations in 3 variables (we have managed to get rid of a):

\begin{aligned} by(y-x) &= 49-7x &-(5) \\  by^2(y-x) &= 133 - 49x &-(6) \\  by^3(y-x) &= 406 - 133x. &-(7) \end{aligned}

The next variable that seems easiest to get rid of is b, and we can do so by performing (6) \div (5) and (7) \div (6):

\begin{aligned} y &= \frac{133-49x}{49-7x} = \frac{19-7x}{7-x} &-(8) \\  y &= \frac{406 - 133x}{133-49x} = \frac{58-19x}{19-7x}. &-(9) \end{aligned}

(We don’t have to worry about dividing by zero as the quantities b, y and y-x all cannot be equal to zero. Why?) We now have 2 equations in 2 variables. The way equations (8) and (9) are set up, it should almost be second nature to continue with the following steps:

\begin{aligned} &&&\frac{19-7x}{7-x} &&= \frac{58-19x}{19-7x} \\  &\Rightarrow &&(19-7x)^2 &&= (58-19x)(7-x) \\  &\Rightarrow &&361 - 266x + 49x^2 &&= 406 - 191x + 19x^2 \\  &\Rightarrow &&30x^2 - 75x - 45 &&= 0 \\  &\Rightarrow &&2x^2 - 5x -3 &&= 0 \\  &\Rightarrow &&(2x+1)(x-3) &&= 0\\  &\Rightarrow &&x &&= -\frac{1}{2} \text{ or } 3.\end{aligned}

Substituting back to get the value of y, we find that (x,y) = (3 -\frac{1}{2}) \text{ or } (-\frac{1}{2}, 3). (This symmetry should not be unexpected: If we look at the original equations, swapping a and x with b and y respectively will result in the same set of equations.) It remains to plug in the values of x and y in the original equations to obtain

(a,b,x,y) = (5, 16, 3, -\frac{1}{2}) \text{ or } (16, 5, -\frac{1}{2}, 3).

It is then a simple matter to evaluate the expression we are asked to evaluate:

\begin{aligned} 2014(x+y-xy) - 100(a+b) &= 2014(3-\frac{1}{2} + \frac{3}{2}) - 100(5+16) \\  &= 2014(4) - 2100 \\  &= 5956. \end{aligned}

The answer is 5956.

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