Problem 37 - Elementary Algebra

Question

For the following problems, solve the linear equations in two variables. $$ 7(t-6)=10(2-s), \text { if } s=5 $$

Step-by-Step Solution

Answer

Answer: The value of \(t\) when \(s=5\) is \(t=\frac{12}{7}\).

Step 1: Write down the given equation

The given equation is: $$ 7(t-6)=10(2-s) $$

Step 2: Plug in the given value of 's'

Now, we will plug in the given value \(s=5\) into the equation: $$ 7(t-6)=10(2-5) $$

Step 3: Simplify the equation

The equation becomes: $$ 7(t-6)=10(-3) $$

Step 4: Solve for 't'

Next, we will isolate 't' on one side of the equation as follows: $$ 7(t-6)=-30 $$ Divide by 7 on both sides: $$ t-6=-\frac{30}{7} $$ Now, add 6 to both sides of the equation: $$ t=-\frac{30}{7}+6 $$ To put the result in the same form (having the same denominator), we can write 6 as \(\frac{42}{7}\): $$ t=-\frac{30}{7}+\frac{42}{7} $$ Then we add both numerators: $$ t=\frac{-30+42}{7} $$ Finally, we simplify the fraction: $$ t=\frac{12}{7} $$

Step 5: State the final solution

Now that we have the value of \(t\), we can state the final solution as: $$ t=\frac{12}{7}, \text { if } s=5 $$

Key Concepts

Linear Equations in Two Variables

Understanding linear equations in two variables is fundamental to algebra. These are equations that involve two different variables, often denoted by letters such as 'x' and 'y' or, as in our exercise, 't' and 's'. The general form of a linear equation in two variables is:
\( ax + by = c \),
where 'a', 'b', and 'c' are constants.Intuitively, this represents a straight line when plotted on a graph with an x and y axis. Every solution to the equation is a point on that line, meaning that there is an infinite number of solutions since a line extends infinitely in both directions. In the context of our exercise:
\( 7(t-6)=10(2-s) \),
when given a specific value for 's', we can find the corresponding value of 't' that satisfies the equation. This single pair of (t,s) is just one of the myriad of points that lie on the line represented by the equation when 's' isn't fixed. By fixing one variable, like setting \( s = 5 \), we reduce the problem to finding a single solution.

Substitution Method

The substitution method is a powerful tool for solving systems of linear equations, which comes particularly handy when we have two equations with two variables. However, it also simplifies solving an equation with already one value given, as in the textbook example. The substitution method involves substituting one variable with its equivalent expression from another equation or, as seen in the example, a given value. Here's how it plays out:

• We start with the original equation \( 7(t-6)=10(2-s) \).
• Since we know that \( s=5 \), we substitute 5 for 's' in the equation.
• We now have an equation with just one variable, 't', which simplifies the solution process.

By making a substitution, we've effectively reduced a two-variable problem to a one-variable problem, which is much more straightforward to solve. This method showcases the power of replacing variables to isolate a single unknown, allowing for easier computation.

Simplifying Equations

Simplifying equations is a critical step in solving algebraic problems efficiently. It involves combining like terms, eliminating unnecessary components, and making the equation as straightforward as possible to solve. The process looks like this:

• Combine like terms on each side of the equation.
• Use the distributive property to eliminate parentheses.
• If fractions are involved, find a common denominator to combine them.
• Carefully perform addition or subtraction to get terms involving the variable on one side and constants on the other.

In our exercise, after substituting 's' with 5, we straightforwardly multiplied and simplified to isolate 't'. We removed the parentheses and simplified the resulting fraction for 't'. Simplifying not only makes the equation easier to manage but also leads to fewer mistakes and clearer solutions. A well-simplified equation lays the groundwork for a successful problem-solving approach, as seen by the clear result obtained in the example where the final simplified expression for 't' is \( \frac{12}{7} \).