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Chapter 10: Problem 39

Find all solutions of the system $$\left\\{\begin{array}{l}x^{3}+y^{3}=3473 \\\x+y=23\end{array}\right.$$

Short Answer

Expert verified
The solutions are \((x, y) = (17, 6)\) and \((6, 17)\).

Step by step solution

01

Express one variable in terms of the other

Start with the equation \( x + y = 23 \). Solve for \( y \) to get an expression: \( y = 23 - x \).
02

Substitute the expression into the cubic equation

Use the expression for \( y \) in terms of \( x \) and substitute it into the cubic equation: \( x^3 + (23-x)^3 = 3473 \).
03

Expand the cubic expression

Expand \((23-x)^3\) using the binomial theorem: \((23-x)^3 = 23^3 - 3 \cdot 23^2 \cdot x + 3 \cdot 23 \cdot x^2 - x^3\). Replace this in the equation: \( x^3 + (23^3 - 3 \cdot 23^2 \cdot x + 3 \cdot 23 \cdot x^2 - x^3) = 3473 \).
04

Simplify the equation

The \( x^3 \) terms cancel out, simplifying to: \( 23^3 - 3 \cdot 23^2 \cdot x + 3 \cdot 23 \cdot x^2 = 3473 \). Calculate \( 23^3 = 12167 \), so we have: \( 12167 - 3(529)x + 69x^2 = 3473 \).
05

Rearrange and solve for x

Rearrange the equation to form a quadratic: \( 69x^2 - 3 \cdot 529 x + (12167 - 3473) = 0 \). Simplify to \( 69x^2 - 1587x + 8694 = 0 \).
06

Use the quadratic formula to find x

Apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 69 \), \( b = -1587 \), and \( c = 8694 \). Calculate the discriminant: \( b^2 - 4ac = 1587^2 - 4 \cdot 69 \cdot 8694\). Solve the quadratic to find \( x \).
07

Calculate the solutions

The discriminant is simplified to give a real, positive value, allowing us to find the square root and calculate the two solutions for \( x \). Substitute these values back into \( y = 23-x \) to find the corresponding \( y \) values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Equations
A cubic equation is any equation that involves a variable raised to the third power. Most commonly, it appears as something like this: \( ax^3 + bx^2 + cx + d = 0 \). Solving cubic equations is like solving puzzles where we aim to find values of variables that satisfy the equation.
  • The degree of a cubic equation is 3, making them more complex compared to quadratic or linear equations.
  • Cubic equations often have two or three real roots, which are the solutions we are interested in.
In the problem we face here, we have a system with a cubic equation, \( x^3 + y^3 = 3473 \), combined with the linear equation \( x + y = 23 \). Together, these allow us to find specific values for both \( x \) and \( y \) that fit both equations.
Substitution Method
The substitution method is a straightforward technique used to solve systems of equations. We solve one equation for one variable and then substitute this expression into the other equation.
  • First, choose an equation that is easily rearranged. Here, we take \( x + y = 23 \) and solve for \( y \), resulting in \( y = 23 - x \).
  • Substitute \( y = 23 - x \) into the cubic equation: \( x^3 + (23-x)^3 = 3473 \).
This method simplifies solving because it reduces the system from two equations in two variables to a single equation with one variable, which can be solved step-by-step.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of quadratic equations鈥攅quations of the form \( ax^2 + bx + c = 0 \). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
  • The expression under the square root sign, \( b^2 - 4ac \), is called the discriminant. It tells us whether real and distinct (positive discriminant), real and repeated (zero discriminant), or complex (negative discriminant) roots exist.
  • In solving our problem, once we have reduced the cubic equation to a quadratic form, we use this formula to find \( x \).
By considering the discriminant's value, we determine the nature of the solutions, which is crucial in verifying their reality and validity.
Binomial Theorem
The binomial theorem provides an efficient way to expand expressions that are raised to a power. It states that: \( (a + b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k \).
  • In the equation \( (23-x)^3 \), the theorem allows us to expand it to \( 23^3 - 3 \cdot 23^2 \cdot x + 3 \cdot 23 \cdot x^2 - x^3 \).
  • This expansion is crucial for simplifying the expressions when tackling the original cubic equation.
Using the binomial theorem simplifies otherwise complex calculations, making it easier to isolate terms and solve for unknowns in equations like our given problem.

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Most popular questions from this chapter

Use a graphing utility to compute the matrix products. $$\left(\begin{array}{rr} -32 & 14 \\ 27 & 9 \end{array}\right)\left(\begin{array}{rr} 83 & -19 \\ 13 & 41 \end{array}\right)$$

Let \(A=\left(\begin{array}{rrr}1 & -6 & 3 \\ 2 & -7 & 3 \\ 4 & -12 & 5\end{array}\right)\) (a) Compute the matrix product \(A A .\) What do you observe? (b) Use the result in part (a) to solve the following system. $$ \left\\{\begin{aligned} x-6 y+3 z &=19 / 2 \\ 2 x-7 y+3 z &=11 \\ 4 x-12 y+5 z &=19 \end{aligned}\right. $$

Use Cramers rule to solve those systems for which \(D \neq 0 .\) In cases where \(D=0,\) use Gaussian elimination or matrix methods. $$\left\\{\begin{array}{l} 5 x-3 y-z=16 \\ 2 x+y-3 z=5 \\ 3 x-2 y+2 z=5 \end{array}\right.$$

The trace of a \(2 \times 2\) matrix \(\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) is defined by \(\operatorname{tr}\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=a+d\) (a) If \(A=\left(\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right)\) and \(B=\left(\begin{array}{ll}5 & 6 \\ 7 & 8\end{array}\right),\) verify that \(\operatorname{tr}(A+B)=\operatorname{tr} A+\operatorname{tr} B\) (b) If \(A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\) and \(B=\left(\begin{array}{ll}e & f \\ g & h\end{array}\right),\) show that \(\operatorname{tr}(A+B)=\operatorname{tr} A+\operatorname{tr} B\).

Solve the following system for \(x\) and \(y\) in terms of \(a\) and \(b\) where \(a b \neq-1\) \(\left\\{\begin{array}{ll}\frac{x+y-1}{x-y+1}=a & \\\ \frac{y-x+1}{x-y+1}=a b & \text { Answer: } x=(a+1) /(a b+1) \\ y=a(b+1) /(a b+1)\end{array}\right.\)
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