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Chapter 5: Problem 10

Determine if the ordered pair is a solution to the system of equations. (See Example 1\()\) \(y=-\frac{1}{5} x+2\) \(2 x+10 y=10\) a. (5,1) b. (-10,4)

Short Answer

Expert verified
Neither (5, 1) nor (-10, 4) is a solution to the system of equations.

Step by step solution

01

Substitute into the First Equation

Substitute the values of the ordered pair \(x, y\) into the first equation \(y = -\frac{1}{5} x + 2\).For a. (5, 1): 1 = -\frac{1}{5} (5) + 2\1 = -1 + 2\1 = 1, which is true.For b. (-10, 4):4 = -\frac{1}{5} (-10) + 2\4 = 2 + 2\4 = 4, which is true.
02

Substitute into the Second Equation

Substitute the values of the ordered pair \(x, y\) into the second equation \(2x + 10y = 10\).For a. (5, 1):2(5) + 10(1) = 10\10 + 10 = 10\20 鈮� 10, which is false.For b. (-10, 4):2(-10) + 10(4) = 10\-20 + 40 = 10\20 = 10, which is false.
03

Conclusion

Since both equations need to be satisfied for the ordered pair to be a solution,For (5, 1): It only satisfies the first equation. Therefore, (5, 1) is not a solution.For (-10, 4): It satisfies the first equation but not the second one. Therefore, (-10, 4) is not a solution.

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Key Concepts

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

Ordered Pairs
In a system of equations, an ordered pair \((x, y)\) represents a potential solution to the set of equations. This pair is considered a solution if it satisfies both equations when the values of \({x}\) and \({y}\) are substituted into each equation. Ordered pairs are expressed in the format \((x, y)\), where the first value is \({x}\) and the second is \({y}\). To determine if the pair is a solution, each equation in the system must hold true when these values are plugged in.
Substitution Method
The substitution method is a technique used to solve systems of equations. This method involves solving one equation for one variable (such as \({y}\)), and then substituting that expression into the other equation.
For example, in the system given:
\[y = -\frac{1}{5} x + 2\]
\[2x + 10y = 10\]
We substitute the ordered pair values into each equation to check if both hold true. Let's consider the pair \( (5, 1) \):
First, substitute \( {x} \) into the first equation:
\[1 = -\frac{1}{5}(5) + 2\]
This simplifies to:
\[1 = -1 + 2\]
Since \[1 = 1\], the first equation is satisfied for this pair. The same process applies to the second equation using the same \((x, y)\) values.
Linear Equations
Linear equations form straight lines when graphed and typically take the form \({ax + by = c}\). In our problem, the system consists of two linear equations:
\[ y = -\frac{1}{5} x + 2\]
\[2x + 10y = 10\]
Each equation represents a line. The solution to the system is the point where these two lines intersect, or, if no intersection point exists that satisfies both equations, then the system has no solution. To find the solution through substitution, we substitute the \( {x} \) and \({y}\) values of given ordered pairs into both equations and check if both equations are true.
Solution Verification
Solution verification involves proving that the values of the ordered pair make both equations in the system true. By substituting \({x}\) and \({y}\) from the ordered pair into each equation, we test for equalities that hold.
For example, to verify whether \( (5, 1) \) is a solution:
  • Substitute \( {x = 5} \) and \( {y = 1} \) into the first equation: \[1 = -\frac{1}{5}(5) + 2\]. It simplifies to \[1 = 1\], which is true.
  • Next, substitute \( {x = 5} \) and \( {y = 1} \) into the second equation: \[2(5) + 10(1) = 10\]. This simplifies to \[20 鈮� 10\], which is false.
Since both equations must be satisfied, \((5, 1)\) is not a solution. This verification process confirms if a pair truly solves the system.

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Most popular questions from this chapter

The attending physician in an emergency room treats an unconscious patient suspected of a drug overdose. The physician does not know the initial concentration \(A_{0}\) of the drug in the bloodstream at the time of injection. However, the physician knows that after \(3 \mathrm{hr}\), the drug concentration in the blood is \(0.69 \mu \mathrm{g} / \mathrm{dL}\) and after \(4 \mathrm{hr}\), the concentration is \(0.655 \mu \mathrm{g} / \mathrm{dL}\). The model \(A(t)=A_{0} e^{-k t}\) represents the drug concentration \(A(t)\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) in the bloodstream \(t\) hours after injection. The value of \(k\) is a constant related to the rate at which the drug is removed by the body. a. Substitute 0.69 for \(A(t)\) and 3 for \(t\) in the model and write the resulting equation. b. Substitute 0.655 for \(A(t)\) and 4 for \(t\) in the model and write the resulting equation. c. Use the system of equations from parts (a) and (b) to solve for \(k .\) Round to 3 decimal places. d. Use the system of equations from parts (a) and (b) to approximate the initial concentration \(A_{0}\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) at the time of injection. Round to 2 decimal places. e. Determine the concentration of the drug after \(12 \mathrm{hr}\). Round to 2 decimal places.

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