Problem 121
Question
Will help you prepare for the material covered in the next section. $$ \text { Solve for } y: \quad x-\frac{5}{y}+4 $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(y = \frac{5}{x-4}\)
1Step 1: Isolate the fraction
Firstly, move all terms not containing \(y\) to the other side of the equation: \(x - 4 = \frac{5}{y}\)
2Step 2: Clear the fraction
To clear out the fraction, multiply all terms by \(y\): \(y(x - 4) = 5\)
3Step 3: Solve for y
Finally, solve for \(y\) by dividing all terms by \((x-4)\): \(y = \frac{5}{x-4}\)
Key Concepts
Isolate VariableClearing FractionsAlgebraic Equations
Isolate Variable
When solving algebraic equations, one of the fundamental steps is to isolate the variable you're solving for. In this case, we are solving for the variable 'y'. The goal here is to get 'y' by itself on one side of the equation, so that it is isolated. This means we will perform operations that will move all other terms to the opposite side.
In our exercise, we start with the equation \( x - \frac{5}{y} + 4 \). To isolate the variable 'y', we first want to get rid of the extra terms on the same side as \( \frac{5}{y} \). This is accomplished by subtracting 'x' and subtracting '4' from both sides of the equation. The result is \( -4 + x = \frac{5}{y} \). Here, 'y' is present in a fraction, which we will next work to clear.
In our exercise, we start with the equation \( x - \frac{5}{y} + 4 \). To isolate the variable 'y', we first want to get rid of the extra terms on the same side as \( \frac{5}{y} \). This is accomplished by subtracting 'x' and subtracting '4' from both sides of the equation. The result is \( -4 + x = \frac{5}{y} \). Here, 'y' is present in a fraction, which we will next work to clear.
Clearing Fractions
The presence of fractions in an algebraic equation can often be an obstacle, but 'clearing' these fractions simplifies the equation and makes it more manageable. Clearing fractions involves eliminating them by using the least common multiple (LCM) or simply multiplying both sides of the equation by the denominator of the fraction.
In our example, we have the fraction \( \frac{5}{y} \). To 'clear' this fraction, we multiply everything by 'y', the denominator of the fraction. This will leave us with 'y' multiplied by the other side of the equation. In mathematical terms, \( y(x - 4) = 5 \). This step effectively removes the fraction and leaves us with a simpler equation where 'y' is now part of a product that we can further manipulate to isolate it.
In our example, we have the fraction \( \frac{5}{y} \). To 'clear' this fraction, we multiply everything by 'y', the denominator of the fraction. This will leave us with 'y' multiplied by the other side of the equation. In mathematical terms, \( y(x - 4) = 5 \). This step effectively removes the fraction and leaves us with a simpler equation where 'y' is now part of a product that we can further manipulate to isolate it.
Algebraic Equations
Algebraic equations are mathematical statements that assert the equality of two expressions, and they often include one or more variables. Our task when solving an algebraic equation is to find the value(s) for the variable(s) that make the equation true.
After clearing the fractions, we have a straightforward equation \( y(x - 4) = 5 \). To solve for 'y', we must isolate it. We achieve this by dividing both sides of the equation by the term \(x - 4\), which is currently being multiplied by 'y'. The resulting equation \( y = \frac{5}{x - 4} \) gives us the value of 'y' in terms of 'x'. This is the solution to our original algebraic equation, where 'y' is now fully isolated and expressed as a function of 'x'.
After clearing the fractions, we have a straightforward equation \( y(x - 4) = 5 \). To solve for 'y', we must isolate it. We achieve this by dividing both sides of the equation by the term \(x - 4\), which is currently being multiplied by 'y'. The resulting equation \( y = \frac{5}{x - 4} \) gives us the value of 'y' in terms of 'x'. This is the solution to our original algebraic equation, where 'y' is now fully isolated and expressed as a function of 'x'.
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