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Chapter 8: Problem 11

Solve the system by the method of substitution. Check your solution graphically. $$\left\\{\begin{array}{c} 2 x+y=6 \\ -x+y=0 \end{array}\right.$$

Short Answer

Expert verified
The solution to the system of equations is \( x = 2, y = 2 \).

Step by step solution

01

Express one variable in terms of the other

From the second equation \( -x + y = 0 \), express \( y \) in terms of \( x \) as follows: \( y = x \).
02

Substitution

Substitute \( y = x \) from step 1 into the first equation \( 2x + y = 6 \). You'll get \( 2x + x = 6 \) which simplifies to \( 3x = 6 \). Solve this to obtain \( x = 2 \). The solution here means when the value of x is 2, both equations in the system are satisfied/solved.
03

Solve for the other variable

Substitute \( x = 2 \) into the equation from step 1 \( y = x \) to get \( y = 2 \). This means when the value of y is 2, both equations in the system are satisfied.
04

Check your solution graphically

To check the solution graphically, plot the two lines represented by the equations \(2x + y = 6\) and \(-x + y = 0\) on the same graph. The point of intersection should be at \( (2,2) \), the solution obtained for the system.

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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 an effective way to solve systems of equations. This technique involves expressing one variable in terms of the other using one of the equations and then substituting this expression into the other equation. Here's how it works in detail for our given example:
  • Step 1: You start with two equations, such as \(-x + y = 0\) and \(2x + y = 6\).
  • Choose one equation to express a variable in terms of the other. Here, from \(-x + y = 0\), we express \(y\) as \(y = x\).
  • Step 2: Substitute this expression (\(y = x\)) into the other equation (\(2x + y = 6\)). This replacement results in one equation with a single variable: \(2x + x = 6\), simplifying to \(3x = 6\).
  • Step 3: Solve for the substituted variable, \(x\). Once solved, we find \(x = 2\).
  • Step 4: Replace \(x\) back into the equation \(y = x\) to solve for \(y\), giving us \(y = 2\).
Through substitution, you narrow down to specific values for each variable that satisfy all original equations.
graphical solution
A graphical solution provides a visual representation of the answers for a system of equations. By plotting the lines of each equation on a coordinate plane, we can observe where these lines intersect, which indicates the solution to the system. In our case:
  • First, consider the equation \(2x + y = 6\).
  • To graph this correctly, rearrange it to \(y = -2x + 6\), marking points like \((0, 6)\) and \((3, 0)\).
  • Next, take the equation \(-x + y = 0\).
  • Rearrange it to become \(y = x\), with points such as \((0, 0)\) and \((2, 2)\).
  • Graph both lines on the same axis. The intersection, in this case, occurs at \((2, 2)\), matching our algebraic solution.
Using a graphical approach, students can verify the solution derived through algebraic methods.
intersection of lines
The intersection of lines concept is pivotal in understanding solutions to systems of equations. In a two-variable system, solving these systems often involves finding the point where their graphical representations intersect:
  • The intersection represents values of variables that satisfy both equations simultaneously.
  • For our equations \(2x + y = 6\) and \(-x + y = 0\), their graphical lines meet at point \((2, 2)\).
  • This point of intersection directly corresponds to the solution found using the substitution method, verifying our results.
  • If lines intersect at one point, it indicates there is a single unique solution to the system.
The understanding of line intersections helps not only in confirming solutions but in enhancing visualization of algebraic processes.

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Most popular questions from this chapter

Use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form . Use the matrix capabilities of a graphing utility to write the matrix in reduced row-echelon form. $$\left[\begin{array}{ccc} 1 & 3 & 2 \\ 5 & 15 & 9 \\ 2 & 6 & 10 \end{array}\right]$$

(A) find the determinant of \(A,\) (b) find \(A^{-1},\) (c) find \(\operatorname{det}\left(A^{-1}\right),\) and (d) compare your results from parts (a) and (c). Make a conjecture based on your results. $$A=\left[\begin{array}{rrr} -1 & 3 & 2 \\ 1 & 3 & -1 \\ 1 & 1 & -2 \end{array}\right]$$

Write an equation of the line passing through the two points. Use the slope- intercept form, if possible. If not possible, explain why. $$(3,4),(10,6)$$

A florist is creating 10 centerpieces for the tables at a wedding reception. Roses cost \(\$ 2.50\) each, lilies cost \(\$ 4\) each, and irises cost \(\$ 2\) each. The customer has a budget of \(\$ 300\) for the centerpieces and wants each centerpiece to contain 12 flowers, with twice as many roses as the number of irises and lilies combined. (a) Write a linear system that represents the situation. (b) Write a matrix equation that corresponds to your system. (c) Solve your linear system using an inverse matrix. Find the number of flowers of each type that the florist can use to create the 10 centerpieces.

Evaluate the determinant, in which the entries are functions. Determinants of this type occur when changes of variables are made in calculus. $$\left|\begin{array}{cc} 4 u & -1 \\ -1 & 2 v \end{array}\right|$$
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