Problem 11
Question
Solve the system by the method of substitution. Check your solution graphically. $$\left\\{\begin{array}{c} 2 x+y=6 \\ -x+y=0 \end{array}\right.$$
Step-by-Step Solution
Verified Answer
The solution to the system of equations is \( x = 2, y = 2 \).
1Step 1: Express one variable in terms of the other
From the second equation \( -x + y = 0 \), express \( y \) in terms of \( x \) as follows: \( y = x \).
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
Key Concepts
substitution methodgraphical solutionintersection of lines
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 \(y = 2\).
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.
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.
Other exercises in this chapter
Problem 11
Find the inverse of the matrix (if it exists). $$\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right]$$
View solution Problem 11
Determine the dimension of the matrix. $$\left[\begin{array}{rr} 33 & 45 \\ -9 & 20 \end{array}\right]$$
View solution Problem 11
Solve the system by the method of elimination. Label each line with its equation. $$\left\\{\begin{aligned} x-y &=2 \\ -2 x+2 y &=5 \end{aligned}\right.$$
View solution Problem 11
Equality of Matrices Find \(x\) and \(y\) or \(x, y,\) and \(z.\) $$\left[\begin{array}{rrr} 4 & 5 & 4 \\ 13 & 15 & 3 y \\ 2 & 2 z-6 & 0 \end{array}\right]=\lef
View solution