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Chapter 3: Problem 62

Solve using any algebraic method. $$ \begin{aligned} &a+5 b=3\\\ &b-7=a \end{aligned} $$

Short Answer

Expert verified
\(b = \frac{5}{3}\), \(a = \frac{-16}{3}\)

Step by step solution

01

Substitute the second equation into the first equation

The second equation is: \(b - 7 = a\) Solve for \(a\) in terms of \(b\): \(a = b - 7\) Substitute \(a\) into the first equation: \((b - 7) + 5b = 3\)
02

Simplify the equation

Combine like terms to simplify the equation: \(b - 7 + 5b = 3\) \(6b - 7 = 3\)
03

Solve for \(b\)

Isolate \(b\) on one side of the equation: \(6b - 7 = 3\) Add 7 to both sides: \(6b = 10\) Divide both sides by 6: \(b = \frac{10}{6} = \frac{5}{3}\)
04

Solve for \(a\) using the value of \(b\)

Substitute \(b = \frac{5}{3}\) into the expression for \(a\): \(a = b - 7\) \(a = \frac{5}{3} - 7\) Convert 7 to a fraction with the same denominator: \(7 = \frac{21}{3}\) Subtract the fractions: \(a = \frac{5}{3} - \frac{21}{3} = \frac{5 - 21}{3} = \frac{-16}{3}\)

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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 involves solving one equation for a single variable, then using that solution in another equation. In our example, we had the system:\[ a + 5b = 3 \] and \[ b - 7 = a \]. First, we solve the second equation for \(a\). We get \[ a = b - 7 \]. Next, this expression for \(a\) is substituted into the first equation. This removes \(a\) and leaves us with an equation in terms of \(b\), which makes it easier to solve. Using substitution can often simplify more complex equations.
isolating variables
Isolating variables is a critical step to solve linear equations. The idea is to get one variable alone on one side of the equation so it鈥檚 easier to solve. For example, in \[ 6b - 7 = 3 \], we want to isolate \(b\). To do this, we first add 7 to both sides of the equation:\[ 6b - 7 + 7 = 3 + 7 \]. This simplifies to \[ 6b = 10 \]. Then, we divide both sides by 6: \[ b = \frac{10}{6} = \frac{5}{3} \]. This leaves \(b\) isolated and solved. Isolating the variable is about performing operations that simplify the equation while keeping it balanced.
combining like terms
Combining like terms involves simplifying equations by summing similar variables. In our example,\[ (b - 7) + 5b = 3 \] simplifies first by applying the substitution. We rewrite it as : \[ b + 5b - 7 = 3 \]. Combining \(b\) and \(5b\) results in \(6b\), so we get:\[ 6b - 7 = 3 \]. By combining the \(b\) terms, our equation becomes easier to handle. This simplification allows us to isolate the variable more effectively.
fraction operations
Handling fractions is often part of solving equations. In our problem, we solved for \(b\) and got:\[ b = \frac{5}{3} \]. To find \(a\), given \( a = b - 7 \), we substitute \( b = \frac{5}{3} \):\[ a = \frac{5}{3} - 7 \]. We need to convert 7 to \( \frac{21}{3} \) to subtract these fractions correctly. So, we rewrite and calculate: \[ \frac{5}{3} - \frac{21}{3} = \frac{5 - 21}{3} = \frac{-16}{3} \]. Understanding how to convert and manipulate fractions makes solving such equations straightforward.

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$$ \begin{aligned} y+z &=1 \\ x+y+z &=1 \\ x+2 y+2 z &=2 \end{aligned} $$
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