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Chapter 7: Problem 14

Solving a System by Substitution In Exercises \(7-14,\) solve the system by the method of substitution. Check your solution(s) graphically. $$\left\\{\begin{array}{l}{y=x^{3}-3 x^{2}+1} \\ {y=x^{2}-3 x+1}\end{array}\right.$$

Short Answer

Expert verified
The solutions of the given system of equations using the method of substitution are (0,1), (1,-1), and (3,-1).

Step by step solution

01

Substituting the value of y

Considering the two equations, substitute y from the second equation into the first. It results in the equation \(x^{3} - 3x^{2} + 1 = x^{2} - 3x + 1\).
02

Simplify the Resulting Equation

Rearrange the equation obtained in step one to bring all the terms to one side of the equation. Subtract \(x^{2} - 3x + 1\) from both sides and simplify. The simplified equation then becomes \(x^{3} - 3x^{2} - x^{2} + 3x = 0\), which further simplifies to \(x^{3} - 4x^{2} + 3x = 0\).
03

Solve for x

The simplified equation from the previous step is a cubic equation which can be further simplified by factoring out an x, giving \(x*(x^{2} - 4x + 3) = 0\). The quadratic equation \(x^{2} - 4x + 3\)= 0 can be solved by factoring or by using Quadratic formula, which gives \(x = 1\) or \(x = 3\). So the solutions of x are \(x = 0\), \(x = 1\), or \(x = 3\).
04

Find correspondive y values.

Substitute the obtained x values into the original y equation \(y = x^{2} - 3x + 1\), find the corresponding y values for each x solution. For \(x = 0\), \(y = 1\); For \(x = 1\), \(y = -1\); and for \(x = 3\), \(y = -1\). Hence, the solutions for the system are (0,1), (1,-1), and (3,-1).
05

Confirming solutions graphically

Plot both equations on a graph and observe their points of intersection. Notice there will be intersections at the points (0,1), (1,-1), and (3,-1), confirming our algebraic results

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Key Concepts

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

Substitution Method
The substitution method is a powerful technique used to solve systems of equations. It involves replacing one variable with an expression involving the other variable. This allows you to solve the system by reducing it to a single equation with one variable.

Here's how it works:
  • Start with two equations in two variables.
  • Solve one of the equations for one variable in terms of the other.
  • Substitute this expression into the second equation.
  • Solve the resulting single-variable equation.
  • Use the solution to find the value of the other variable.
This method is particularly useful when one of the equations is easily solvable for one of the variables. It simplifies the system, making it easier to find a solution.
Cubic Equations
Cubic equations are polynomial equations of degree three. An example of a cubic equation is one of the form:
  • \[ ax^3 + bx^2 + cx + d = 0 \]
Here, the highest power of the variable is three.

Solving cubic equations can be more complex than linear or quadratic equations. Common methods include factoring, synthetic division, and using the cubic formula. In our problem, the equation was simplified by factoring, which reduced a cubic equation to a lower degree equation.
  • The original form presented was \[ x^3 - 4x^2 + 3x = 0 \], solved by factoring out \( x \).
  • This allowed the remaining quadratic \( x^2 - 4x + 3 = 0 \) to be solved more straightforwardly.
Factors are the key in simplifying cubics when it allows breaking down the polynomial.
Quadratic Equations
Quadratic equations are polynomial equations of degree two, and they are written in the standard form:
  • \[ ax^2 + bx + c = 0 \]
Solving quadratic equations can be done using various methods, such as:
  • Factoring
  • Completing the square
  • The quadratic formula \[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
In the exercise, the quadratic equation emerged after factoring out the variable in the cubic equation. The obtained quadratic, \( x^2 - 4x + 3 \), was easily solved by simple factoring, leading to solutions for \( x \) that included the points of intersection for our original system.
Graphical Solutions
Graphical solutions involve plotting each equation on a coordinate plane and identifying where they intersect. This is an excellent way to verify solutions obtained algebraically.

For our system of equations:
  • The first equation is \( y = x^3 - 3x^2 + 1 \)
  • The second equation is \( y = x^2 - 3x + 1 \)
By plotting these equations, you can visually confirm the solutions derived algebraically. The intersections of the graphs represent the common solutions to the equations.

This method not only checks accuracy but also provides a visual understanding of how the solutions relate geometrically. It is a powerful tool in understanding the behavior and relationships of equations within a system.

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Most popular questions from this chapter

Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=5 x+4 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {2 x+2 y \geq 10} \\ {x+2 y \geq 6}\end{array} $$

Describing an Unusual Characteristic, the linear programming problem has an unusual characteristic. Sketch a graph of the solution region for the problem and describe the unusual characteristic. Find the minimum and maximum values of the objective function (if possible) and where they occur. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=3 x+4 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {x+y \leq 1} \\ {2 x+y \leq 4}\end{array} $$

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