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 \(c\).)



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 \((x,y,z)\).)

 Nov 1, 2023

(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}

 Nov 1, 2023


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.

 Nov 1, 2023

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