+0

0
5
3
+23

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 $$(x,y,z)$$.)

Nov 1, 2023

#1
+136
0

(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
#2
+136
0

(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.

Nov 1, 2023