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Chapter 10: Problem 21

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered pair form given in Example 6. $$\left\\{\begin{array}{r}x+y=4 \\\\-x+y=0\end{array}\right.$$

Short Answer

Expert verified
The unique solution is \((2, 2)\).

Step by step solution

01

Understand the System of Equations

The given system of equations is: \[ \begin{cases} x + y = 4 & \text{(Equation 1)} \ -x + y = 0 & \text{(Equation 2)} \end{cases} \] Our goal is to find the values of \( x \) and \( y \) that satisfy both equations.
02

Solve Equation 2 for y

Equation 2 is \(-x + y = 0\). Solve for \(y\): \[ y = x \] This expresses \(y\) in terms of \(x\).
03

Substitute y in Equation 1

Substitute \(y = x\) from Step 2 into Equation 1:\[ x + (x) = 4 \]This simplifies to \[ 2x = 4 \].
04

Solve for x

Solve the equation \(2x = 4\) for \(x\):\[ x = \frac{4}{2} \]\[ x = 2 \]
05

Find y using x=2

Use the expression for \(y\) from Step 2, \(y = x\), and substitute the value of \(x = 2\):\[ y = 2 \]
06

Verify the Solution

Verify that \(x = 2\) and \(y = 2\) satisfy both original equations:- For Equation 1: \(2 + 2 = 4\) which is true.- For Equation 2: \(-2 + 2 = 0\) which is true.Both equations are satisfied, confirming the solution is correct.

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Key Concepts

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

Linear Equations
A linear equation is a type of equation where the variables are raised only to the power of one. It forms a straight line when graphed on a coordinate plane. In our exercise, we have two linear equations given as:
  • Equation 1: \( x + y = 4 \)
  • Equation 2: \( -x + y = 0 \)
Each equation represents a straight line. To solve for the variables \(x\) and \(y\), we need to find the point where these two lines intersect on the coordinate plane. This point is known as the solution to the system of equations.
Solving Algebraically
Solving algebraically refers to finding the variable values that satisfy the given equations using algebraic manipulations. For the system of equations, it's often easier to solve one of the equations for a single variable and then substitute it into the other equation.
In our example, we first solved Equation 2 for \(y\):
  • \( -x + y = 0 \)
  • Adding \(x\) to both sides gives: \( y = x \)
Now, we have expressed \(y\) in terms of \(x\), which helps us reduce the number of variables in the equations, making it easier to solve the entire system.
Substitution Method
The substitution method is a widely used approach to solve systems of equations. It involves isolating one variable in one of the equations and then substituting it into the other equation. This method simplifies the system to a single equation with one variable.
In the given problem, using the substitution method involved substituting \( y = x \) (from our solution of Equation 2) back into Equation 1:
  • Substituting gives: \( x + (x) = 4 \)
  • This equation simplifies to \( 2x = 4 \)
  • Divide both sides by 2 to solve for \(x\): \( x = 2 \)
Once we found \(x\), we used the initial substitution \( y = x \) to find \( y = 2 \). The substitution method is powerful because it simplifies complex systems into smaller, more manageable problems.

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Most popular questions from this chapter

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{array}{c} x+2 y \leq 14 \\ 3 x-y \geq 0 \\ x-y \geq 2 \end{array}\right.$$

Use a calculator that can perform matrix operations to solve the system, as in Example 7 . $$\left\\{\begin{array}{l} 3 x+4 y-z=2 \\ 2 x-3 y+z=-5 \\ 5 x-2 y+2 z=-3 \end{array}\right.$$

Solve the system of equations by converting to a matrix equation and using the inverse of the coefficient matrix, as in Example 6 . Use the inverses from Exercises 9-12,17,18,21, and 23. $$\left\\{\begin{array}{l} 3 x+4 y=10 \\ 7 x+9 y=20 \end{array}\right.$$

Polynomials Determined by a Set of Points We all know that two points uniquely determine a line \(y=a x+b\) in the coordinate plane. Similarly, three points uniquely determine a quadratic (second-degree) polynomial $$y=a x^{2}+b x+c$$ four points uniquely determine a cubic (third-degree) polynomial $$y=a x^{3}+b x^{2}+c x+d$$ and so on. (Some exceptions to this rule are if the three points actually lie on a line, or the four points lie on a quadratic or line, and so on.) For the following set of five points, find the line that contains the first two points, the quadratic that contains the first three points, the cubic that contains the first four points, and the fourth-degree polynomial that contains all five points. $$(0,0), \quad(1,12), \quad(2,40), \quad(3,6), \quad(-1,-14)$$ Graph the points and functions in the same viewing rectangle using a graphing device.

Find the inverse of the matrix. For what value(s) of \(x\) if any, does the matrix have no inverse? $$\left[\begin{array}{ll} 2 & x \\ x & x^{2} \end{array}\right]$$
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