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Chapter 12: Problem 15

Solve each system using the substitution method. \(y=x^{2}+6 x+9\) \(x+y=3\)

Short Answer

Expert verified
The solutions are (-1, 4) and (-6, 9).

Step by step solution

01

Express One Variable in Terms of Another

From the second equation, express y in terms of x. Given: \[ x + y = 3 \] Rearranging this equation, we have: \[ y = 3 - x \]
02

Substitute the Expression into the First Equation

Substitute \( y = 3 - x \) into the first equation. Given: \[ y = x^2 + 6x + 9 \] Substitute \( y \): \[ 3 - x = x^2 + 6x + 9 \]
03

Solve the Resulting Quadratic Equation

To solve the quadratic equation, first move all terms to one side:\[ x^2 + 6x + 9 + x - 3 = 0 \] Simplify: \[ x^2 + 7x + 6 = 0 \] Factorize the quadratic equation: \[ (x + 1)(x + 6) = 0 \] Set each factor to zero: \[ x + 1 = 0 \] \[ x + 6 = 0 \] Solve for x: \[ x = -1 \] \[ x = -6 \]
04

Find Corresponding y Values

Substitute the x-values back into the equation \( y = 3 - x \): For \( x = -1 \): \[ y = 3 - (-1) = 4 \] For \( x = -6 \): \[ y = 3 - (-6) = 9 \]
05

Write the Solutions as Ordered Pairs

The solutions are the ordered pairs of x and y values found: \[ (-1, 4) \] \[ (-6, 9) \]

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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 popular way to solve a system of equations. The idea is simple: solve one of the equations for one variable in terms of the other variable. Then, substitute that expression into the other equation. This turns a system of equations into a single equation, making it easier to solve. In our example, we had the system:

  • \(y = x^2 + 6x + 9\)
  • \(x + y = 3\)
First, we solve the second equation for y:

\[ y = 3 - x \]

Next, we substitute \(3 - x\) for y in the first equation. This helps us solve for x. Once we find x, we can easily find the corresponding y values.
solving quadratic equations
Quadratic equations usually have the form \(ax^2 + bx + c = 0\). Solving these involves finding the values of x that make the equation true. These values are called 'roots' or 'solutions'. Let's see how we solve them using our problem. After substituting \(y = 3 - x\) into \(y = x^2 + 6x + 9\), we got:

\[ 3 - x = x^2 + 6x + 9 \]

By rearranging and combining like terms, the equation became:

\[ x^2 + 7x + 6 = 0 \]
factoring quadratics
Factoring is a method to solve quadratic equations. We express the quadratic as a product of simpler expressions set to zero, making it easier to find the roots. For \(x^2 + 7x + 6 = 0\), we factor it into:

\[ (x + 1)(x + 6) = 0 \]

By setting each factor to zero:\( x + 1 = 0 \) and \( x + 6 = 0 \), we find:

  • \(x = -1\)
  • \(x = -6\)
These are the x-values for our solutions.
ordered pairs
Ordered pairs help us represent solutions to systems of equations. Each pair includes a value for x and a corresponding value for y. From our example, the x-values we found were \(-1\) and \(-6\). To find y-values, substitute these x-values back into \(y = 3 - x\):

  • For \(x = -1:\) \(y = 3 - (-1) = 4\)
  • For \(x = -6:\) \(y = 3 - (-6) = 9\)
Therefore, the ordered pairs (solutions) are:

  • \((-1, 4)\)
  • \((-6, 9)\)

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Most popular questions from this chapter

Solve each system using the elimination method or a combination of the elimination and substitution methods. $$ \begin{array}{c} x^{2}+3 y^{2}=40 \\ 4 x^{2}-y^{2}=4 \end{array} $$

Solve each system using the elimination method or a combination of the elimination and substitution methods. $$ \begin{array}{r} -2 x^{2}+7 x y-3 y^{2}=4 \\ 2 x^{2}-3 x y+3 y^{2}=4 \end{array} $$

Solve each system using the substitution method. \(y=x^{2}+8 x+16\) \(x-y=-4\)

Graph each inequality. \(y^{2}-16 x^{2} \leq 16\)

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