Problem 15
Question
Solve the equation and describe each step you use. $$ 7(1-y)=-3(y-2) $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(y = \frac{1}{4}\).
1Step 1: Simplify both sides
Firstly, proceed to simplify both sides of the equation. Distribute 7 across (1-y) on the left side to obtain \(7 - 7y\), and distribute -3 across (y-2) on the right side to obtain \(-3y + 6\). Now, our equation is \(7 - 7y = -3y + 6\).
2Step 2: Gather like terms
Proceed to gather like terms on both sides. To do this, add 3y to both sides of the equation to remove -3y on the right side and add 7y to the right side to eliminate -7y from the left side. This results in \[ 7 = 4y+6 \].
3Step 3: Determine y value
Now that we've gathered like terms, we need to isolate y. Subtract 6 from both sides to remove 6 from the right side of the equation. This will give us \[ 1 = 4y \]. Then, divide by 4 on both sides to isolate y. The result is \[ y = \frac{1}{4} \].
Key Concepts
The Distributive PropertyCombining Like TermsIsolating Variables
The Distributive Property
Understanding the distributive property is crucial when tackling linear equations. It's the bedrock that allows us to make equations more manageable and solvable. In essence, the distributive property lets us multiply a single term by each term within a parenthesis.
For instance, you'll often see it in action when you come across an equation like the one from our exercise, where we must distribute the number outside the parentheses across each term inside. Looking at our exercise, we have the equation \(7(1-y)=-3(y-2)\). To apply the distributive property, we multiply 7 by each term inside the left parenthesis and -3 by each term inside the right parenthesis. The result is \(7\times1 - 7\times y = -3\times y + (-3)\times(-2)\), which simplifies to \(7 - 7y = -3y + 6\).
Often, this step helps reveal the 'like terms' that we can then combine to further simplify the equation. It's a methodical approach and, by following it, calculational mistakes can be significantly reduced.
For instance, you'll often see it in action when you come across an equation like the one from our exercise, where we must distribute the number outside the parentheses across each term inside. Looking at our exercise, we have the equation \(7(1-y)=-3(y-2)\). To apply the distributive property, we multiply 7 by each term inside the left parenthesis and -3 by each term inside the right parenthesis. The result is \(7\times1 - 7\times y = -3\times y + (-3)\times(-2)\), which simplifies to \(7 - 7y = -3y + 6\).
Often, this step helps reveal the 'like terms' that we can then combine to further simplify the equation. It's a methodical approach and, by following it, calculational mistakes can be significantly reduced.
Combining Like Terms
After distributing, we're left with an equation that usually has terms on both sides that are similar in nature, also known as 'like terms.' In the environment of algebra, 'like terms' are terms that have the same variables raised to the same power. The process of combining like terms is essential for simplifying equations and moving toward a solution.
In our worked problem, once we've distributed we have \(7 - 7y = -3y + 6\). Here, \-7y \( and \)-3y are like terms, as they are both coefficients of the variable \(y\). To combine them, you perform the arithmetic operation indicated; in this case, we add \(+3y\) to both sides.
This gives us a new simplified equation: \(7 = 4y + 6\). With this simplification, you've effectively set the stage for the final act of the algebraic play - isolating the variable to find its value.
In our worked problem, once we've distributed we have \(7 - 7y = -3y + 6\). Here, \-7y \( and \)-3y are like terms, as they are both coefficients of the variable \(y\). To combine them, you perform the arithmetic operation indicated; in this case, we add \(+3y\) to both sides.
This gives us a new simplified equation: \(7 = 4y + 6\). With this simplification, you've effectively set the stage for the final act of the algebraic play - isolating the variable to find its value.
Isolating Variables
The grand finale of solving linear equations involves isolating the variable you're solving for, which means getting the variable by itself on one side of the equation and everything else on the other side. This step is where we really get to the heart of the matter and find the solution we've been searching for.
Let's take our simplified equation \(7 = 4y + 6\). To isolate \(y\), we want to get rid of the constant term \(+6\) that's on the same side as the variable. We do this by doing the opposite operation—subtraction in our case. So we subtract 6 from both sides and obtain \(1 = 4y\).
Now, \(y\) is the only term on its side, but it still has a coefficient of 4. We want \(y\) alone, so we divide both sides of the equation by 4, leaving us with \(y = \frac{1}{4}\). There you have it, \(y\) is completely isolated, and we've discovered its value in this equation.
Let's take our simplified equation \(7 = 4y + 6\). To isolate \(y\), we want to get rid of the constant term \(+6\) that's on the same side as the variable. We do this by doing the opposite operation—subtraction in our case. So we subtract 6 from both sides and obtain \(1 = 4y\).
Now, \(y\) is the only term on its side, but it still has a coefficient of 4. We want \(y\) alone, so we divide both sides of the equation by 4, leaving us with \(y = \frac{1}{4}\). There you have it, \(y\) is completely isolated, and we've discovered its value in this equation.
Other exercises in this chapter
Problem 15
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