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Chapter 3: Problem 19

Solve each system by substitution. See Examples 1 and 2 . $$ \left\\{\begin{array}{l} 0.3 a+0.1 b=0.5 \\ \frac{4}{3} a+\frac{1}{3} b=3 \end{array}\right. $$

Short Answer

Expert verified
The solution is \((a, b) = (4, -7)\).

Step by step solution

01

Solve One Equation for One Variable

Start with the first equation: \(0.3a + 0.1b = 0.5\). Let's solve for \(b\) in terms of \(a\). First, isolate \(b\): \(0.1b = 0.5 - 0.3a\). Then, solve for \(b\) by dividing through by 0.1: \(b = \frac{0.5 - 0.3a}{0.1}\). Simplify the expression: \(b = 5 - 3a\).
02

Substitute into the Second Equation

Substitute \(b = 5 - 3a\) from Step 1 into the second equation: \(\frac{4}{3}a + \frac{1}{3}(5 - 3a) = 3\).
03

Simplify and Solve for 'a'

Distribute and simplify the second equation: \(\frac{4}{3}a + \frac{5}{3} - a = 3\). Combine like terms: \(\left(\frac{4}{3} - 1\right)a + \frac{5}{3} = 3\) which simplifies to \(\frac{1}{3}a + \frac{5}{3} = 3\). Subtract \(\frac{5}{3}\) from both sides: \(\frac{1}{3}a = 3 - \frac{5}{3}\). Simplify the right-hand side: \(\frac{1}{3}a = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}\). Multiply by 3 to solve for \(a\), resulting in \(a = 4\).
04

Replace 'a' in the Expression for 'b'

Using the expression \(b = 5 - 3a\) from Step 1, substitute \(a = 4\): \(b = 5 - 3(4)\). Calculate \(b = 5 - 12 = -7\).
05

Check Your Solution

Verify the solution \((a, b) = (4, -7)\) by substituting back into the original equations: For the first equation, \(0.3(4) + 0.1(-7) = 1.2 - 0.7 = 0.5\), which is correct. For the second equation, \(\frac{4}{3}(4) + \frac{1}{3}(-7) = \frac{16}{3} - \frac{7}{3} = 3\), which is also correct. The solutions satisfy both equations.

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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 powerful and straightforward technique for solving systems of equations. Let's explore how it works using our two linear equations:
  • Step 1 is all about picking one equation and solving it for one of the variables. For instance, we changed the first equation to express \( b \) in terms of \( a \): \( b = 5 - 3a \). This step involves isolating the variable on one side of the equation, making it easier to substitute into the other equation.
  • In Step 2, we substitute the expression we found for \( b \) into the second equation. This means replacing every instance of \( b \) in that equation with \( 5 - 3a \), which simplifies the task significantly, reducing the number of variables to manage.
Using the substitution method simplifies the complexity of dealing with two variables at once. It's especially useful when equations are already set to allow one variable to be straightforwardly isolated.
Linear Equations
Linear equations are the foundation of algebra and model a relationship in which each variable corresponds to a constant rate of change. In the system given:
  • Each equation represents a straight line in a coordinate system, with solutions being the points where these lines intersect.
  • The equation \( 0.3a + 0.1b = 0.5 \) represents one line, and \( \frac{4}{3}a + \frac{1}{3}b = 3 \) represents another. Where they meet is the solution to the system, indicating values for \( a \) and \( b \) that satisfy both equations simultaneously.
Understanding how linear equations work not only helps in solving systems but also in visualizing how equations translate graphically. When plotted, linear equations are straight lines, and their intersection point provides valuable insights into real-world scenarios modelled by algebra.
Algebraic Manipulation
Algebraic manipulation is a set of techniques used to rearrange equations and expressions, making them easier to work with. In our solution:
  • We used basic rules of algebra to isolate variables and simplify expressions. For example, solving \( 0.1b = 0.5 - 0.3a \) involved dividing by 0.1 to get \( b = 5 - 3a \).
  • Another crucial step was distributing the \( \frac{1}{3} \) in the equation \( \frac{1}{3}(5 - 3a) \), leading to further simplification.
  • Further, combining like terms, such as in \( \left(\frac{4}{3} - 1\right)a + \frac{5}{3} = 3 \), helped to streamline the process.
These algebraic maneuvers simplify complex equations and are central to solving them efficiently. The ability to manipulate equations using these rules is crucial not just in algebra but across all areas of mathematics.

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Most popular questions from this chapter

Snowmobiling. A man rode a snowmobile at the rate of \(20 \mathrm{mph}\) and then skied cross country at the rate of \(4 \mathrm{mph}\). During the 6-hour trip, he traveled 48 miles. How long did he snowmobile, and how long did he ski?

Digital Photography. A digital camera stores the black and white photograph shown on the right as a \(512 \times 512\) matrix. Each element of the matrix corresponds to a small dot of grey scale shading, called a pixel, in the picture. How many elements does a \(512 \times 512\) matrix have?

Use Cramer's rule to solve each system of equations. If a system is inconsistent or if the equations are dependent, so indicate. $$ \left\\{\begin{array}{l} 2 x+3 y=0 \\ 4 x-6 y=-4 \end{array}\right. $$

Explain how to find \(x\) when solving a system of three linear equations in \(x, y,\) and \(z\) by Cramer's rule. Use the words coefficients and constants in your explanation.

How are the graphs of \(f(x)=x^{2}\) and \(g(x)=x^{2}-2\) related?
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