+28 A)
In this multi-part problem, we will consider this system of simultaneous equations: \(\begin{array}{r@{~}c@{~}l l} 3x+5y-6z &=&2, & \textrm{(i)} \\ 5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\ xyz&=&6. & \textrm{(iii)} \end{array}\)
Let \(a=3x\), \(b=5y\), and \(c=-6z\). Determine the monic cubic polynomial in terms of a variable \(t\) whose roots are\(t=a\), \(t=b\), and \(t=c\). Make sure to enter your answer in terms of \(t\)and only \(t\), in expanded form. (Suggestion: As a first step, you may wish to rewrite the system of equations in terms of \(a\),\(b\), and \(c\).)
B)
This is a continuation of the problem above. Given that \((x,y,z)\) is a solution to the original system of equations, determine all distinct possible values of \(x +y\) (Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple \((a,b,c)\), then determine the possible solutions \((x,y,z)\).)
+1533 (A) For any number a, we can substitute a into the cubic equation f(x)=x3+px2+qx+r, and we get: \begin{align*} f(a)&= a^3 + pa^2 + qa + r \ &= a(a^2 + pa + q) + r \end{align*}The term (a2+pa+q) is the quadratic expression that has the roots a and −p. We can also think of it as the sum of the roots of the quadratic expression. So if we want a cubic expression with the roots a, b, and c, then the quadratic expression with the roots a and −p is (a+b+c). We can use the same logic for the other two quadratic expressions that we need to create. We get: \begin{align*} a(a+b+c) + r = a^3 + ab + ac + r\ b(a+b+c) + r = ab^2 + bb + bc + r\ c(a+b+c) + r = ac^2 + bc + cc + r \end{align*}Subtracting the first equation from the second equation, we get: \begin{align*} b(a+b+c) + r - (a(a+b+c) + r) &= 0\ b(a+b+c) - a(a+b+c) &= 0\ (b-a)(a+b+c) &= 0\ (b-a)\left(t+5\right)&=0 \end{align*}We can do the same for the other pair of equations: \begin{align*} (c-a)\left(t+6\right)&=0\ (c-b)\left(t+5\right)&=0 \end{align*}Multiplying these three equations together, we get the monic cubic expression that we want. \begin{align*} &\left(t+5\right)\left(t+6\right)\left(t+5\right) \ &= (t^2+11t+30)(t+5)\ &= t^3+16t^2+65t+150 \end{align*}Therefore, the monic cubic polynomial in terms of a variable t whose roots are t=a, t=b, and t=c is: \boxed{t^3+16t^2+65t+150}
+1533 (B)
We can solve the system of equations for x, y, and z by using the following steps:
Multiply the first equation by −6 and the second equation by 5.
Add the two equations together to get −6xy+30yz+30xz=22.
Divide both sides of the equation by −6.
Simplify the equation to get xy−5yz−5xz=−11.
Add the equation xyz=6 to the equation xy−5yz−5xz=−11 to get 6xy−5yz−5xz=−5.
Factor the equation to get (x−1)(6y−5z)=5.
Since x, y, and z are all non-zero, we must have x−1=5 and 6y−5z=−1.
Solve for x and y to get x=6 and y=65z−1.
Substitute these values into the equation xyz=6 to get 6z=636z−6.
Solve for z to get z=21.
Substitute z=21 into the equation y=65z−1 to get y=31.
Therefore, the only possible value of x+y is 6+1/3 = 19/3.
