Solving Algebraic Equations: (2y+a)/(2y+b) = A/b

When you're diving into the world of algebraic equations, you'll encounter a variety of problems that test your understanding of how variables and constants interact. One such intriguing problem is solving for 'y' in the equation $\frac{2 y+a}{2 y+b}=\frac{a}{b}$. This isn't just about finding a numerical answer; it's about manipulating expressions, understanding the properties of fractions, and being careful about potential pitfalls like division by zero. We'll break down this equation step-by-step, making sure to cover all the bases and ensure you feel confident in your ability to tackle similar problems. Remember, mathematics is a journey of exploration, and each equation solved brings you closer to a deeper comprehension of its underlying principles. So, let's get started on unraveling this algebraic puzzle, and by the end, you'll see that with a clear approach, even seemingly complex fractions can be demystified.

Understanding the Equation

Before we jump into solving, it's crucial to understand the components of the equation $\frac{2 y+a}{2 y+b}=\frac{a}{b}$. We have two fractions set equal to each other. On the left side, the numerator is $2y+a$ and the denominator is $2y+b$. On the right side, we have the ratio of $a$ to $b$. Our goal is to isolate the variable 'y'. However, we must also be mindful of the conditions under which this equation is valid. Specifically, the denominators cannot be zero. This means $2y+b \neq 0$ and $b \neq 0$. If $b=0$, the right side of the equation would be undefined. Also, if $2y+b=0$, the left side would be undefined. These constraints are important because they define the domain of possible solutions for 'y'. In many algebraic problems, identifying these restrictions early on can prevent errors and help you interpret your final answer correctly. Think of these as the 'rules of the game' for this particular equation. Without respecting these rules, any solution we find might be extraneous, meaning it doesn't actually satisfy the original equation.

The Cross-Multiplication Method

One of the most straightforward ways to solve equations involving proportions like this is through cross-multiplication. This technique is based on the fundamental property of equality: if $\frac{p}{q} = \frac{r}{s}$, then $ps = qr$, provided $q \neq 0$ and $s \neq 0$. Applying this to our equation $\frac{2 y+a}{2 y+b}=\frac{a}{b}$, we multiply the numerator of the left fraction by the denominator of the right fraction and set it equal to the product of the denominator of the left fraction and the numerator of the right fraction. So, we get: $(2y+a) \times b = (2y+b) \times a$. This step effectively eliminates the denominators, transforming the fractional equation into a linear equation that is generally much easier to solve. This method is a cornerstone of solving proportions and is widely applicable in various areas of mathematics and science where ratios and proportional relationships are important. It's a powerful tool that simplifies complex fraction setups into more manageable algebraic forms, making the path to finding the unknown variable significantly clearer.

Expanding and Simplifying

Once we have applied cross-multiplication, the next step is to expand both sides of the equation to remove the parentheses and simplify. Applying the distributive property, we get: $2yb + ab = 2ya + ba$. Notice that $ab$ and $ba$ represent the same quantity since multiplication is commutative. Therefore, we can simplify this to: $2yb + ab = 2ya + ab$. Our next move is to gather all terms containing 'y' on one side of the equation and all constant terms (terms without 'y') on the other side. To do this, we can subtract $ab$ from both sides: $2yb = 2ya$. This simplification brings us closer to isolating 'y'. It's essential here to be methodical with each algebraic step. Mistakes can easily creep in if you're not careful with signs or if you misapply the distributive property. This stage is where the structure of the equation starts to reveal the solution. The simplification from $2yb + ab = 2ya + ab$ to $2yb = 2ya$ is a key outcome of careful algebraic manipulation, highlighting the power of cancellation when terms appear on both sides of an equation.

Isolating the Variable 'y'

With the equation simplified to $2yb = 2ya$, our objective is to isolate 'y'. To do this, we need to move all terms involving 'y' to one side and all other terms to the other side. However, in this specific simplified form, we have 'y' on both sides. A common strategy is to bring all 'y' terms to one side. Let's subtract $2ya$ from both sides: $2yb - 2ya = 0$. Now, we can factor out $2y$ from the terms on the left side: $2y(b - a) = 0$. This factored form is very revealing. For the product of two factors to be zero, at least one of the factors must be zero. Therefore, we have two possibilities: either $2y = 0$ or $b - a = 0$. If $2y = 0$, then dividing by 2 gives us $y = 0$. If $b - a = 0$, then $b = a$. This leads us to consider the conditions under which our original equation is valid and what these possibilities mean for the solution.

Considering Special Cases and Conditions

It's crucial to revisit the conditions we established earlier: $2y+b \neq 0$ and $b \neq 0$. Let's examine the potential solutions based on these conditions.

• Case 1: $y = 0$. If $y=0$, let's substitute this back into the original equation: $\frac{2(0)+a}{2(0)+b}=\frac{a}{b}$, which simplifies to $\frac{a}{b}=\frac{a}{b}$. This is true as long as $b \neq 0$ and $2(0)+b \neq 0$ (which means $b \neq 0$). So, $y=0$ is a valid solution provided $b \neq 0$.
• Case 2: $b = a$. If $b = a$, then our original equation becomes $\frac{2 y+a}{2 y+a}=\frac{a}{a}$. The right side simplifies to 1 (provided $a \neq 0$). The left side also simplifies to 1, provided $2y+a \neq 0$. So, if $b=a$ and $a \neq 0$, the equation is true for all values of $y$ such that $2y+a \neq 0$, which means $y \neq -\frac{a}{2}$. However, let's look back at our derivation $2y(b-a)=0$. If $b=a$, then $b-a=0$, and the equation becomes $2y(0)=0$, which is $0=0$. This equation is true for any value of $y$. But we must also consider the initial restrictions. If $b=a$, the original equation is $\frac{2y+a}{2y+a} = \frac{a}{a}$. For this to be defined, we need $a \neq 0$ (so the RHS is defined) and $2y+a \neq 0$ (so the LHS is defined). If $a \neq 0$ and $b=a$, then any value of $y$ such that $y \neq -\frac{a}{2}$ is a solution.

However, if we strictly follow the steps of solving for 'y' assuming $b \neq a$, we arrive at $y=0$. Let's re-examine the step $2yb = 2ya$. If we assume $a \neq b$, we can divide both sides by $2(b-a)$. Let's restart from $2yb = 2ya$. If $a \neq b$, we can rearrange to $2yb - 2ya = 0$, which factors to $2y(b-a) = 0$. If $b-a \neq 0$ (i.e., $b \neq a$), then we can divide by $2(b-a)$, which gives $y = \frac{0}{2(b-a)}$, so $y=0$. This solution is valid provided the initial denominators are not zero. If $y=0$, the denominators are $b$ and $b$. So, we need $b \neq 0$. If $b=0$, the original equation is undefined.

What if we rearrange $2yb = 2ya$ differently? Suppose $a \neq 0$. We can divide both sides by $2a$: $\frac{2yb}{2a} = \frac{2ya}{2a}$, which simplifies to $\frac{yb}{a} = y$. Now, subtract $y$ from both sides: $\frac{yb}{a} - y = 0$. Factor out $y$: $y(\frac{b}{a} - 1) = 0$. This again leads to two possibilities: $y=0$ or $\frac{b}{a} - 1 = 0$. If $\frac{b}{a} - 1 = 0$, then $\frac{b}{a} = 1$, which means $b=a$. This confirms our previous findings.

Therefore, the primary solution we find through algebraic manipulation, assuming $a \neq b$ and $b \neq 0$, is $y=0$. The case where $a=b$ leads to a situation where the equation holds true for a wider range of $y$ values, provided the denominators are non-zero.

Final Solution and Verification

The process of solving algebraic equations involves not just manipulation but also verification. We found that $y=0$ is a potential solution, provided certain conditions are met. Let's recap these conditions and the solution.

• The initial equation: $\frac{2 y+a}{2 y+b}=\frac{a}{b}$
• Restrictions: $b \neq 0$ and $2y+b \neq 0$.
• Steps: Cross-multiplication leads to $(2y+a)b = (2y+b)a$, which simplifies to $2yb + ab = 2ya + ab$. Further simplification gives $2yb = 2ya$.
• Analysis of $2yb = 2ya$:
  • If $a \neq b$, we can rearrange to $2y(b-a) = 0$. Since $b-a \neq 0$, we must have $2y=0$, leading to $y=0$.
  • If $a = b$, the equation becomes $2yb = 2ya$. Since $a=b$, this is $2ya = 2ya$, which is true for all $y$. However, we must also consider the original denominators. If $a=b$, the equation is $\frac{2y+a}{2y+a} = \frac{a}{a}$. For this to be defined, we need $a \neq 0$ and $2y+a \neq 0$. So, if $a=b \neq 0$, the equation holds for all $y$ where $y \neq -\frac{a}{2}$.
• Verification of $y=0$: If $y=0$, the original equation becomes $\frac{a}{b} = \frac{a}{b}$. This is valid as long as $b \neq 0$. If $b=0$, the original equation is undefined.

Therefore, the general solution to the equation $\frac{2 y+a}{2 y+b}=\frac{a}{b}$ is $y=0$, provided that $b \neq 0$ and $a \neq b$.

If $a = b$, the equation holds true for all values of $y$ as long as $a \neq 0$ and $2y+a \neq 0$. This means for $a=b \neq 0$, the solution set is all real numbers $y$ except for $y = -\frac{a}{2}$.

It's always a good practice to check your answers. If we plug $y=0$ back into the original equation, we get $\frac{a}{b} = \frac{a}{b}$, which is correct provided $b \neq 0$. The exploration of algebraic equations like this highlights the importance of carefully considering all conditions and potential edge cases to arrive at a complete and accurate solution. Understanding these nuances is key to mastering mathematical problem-solving.

For further exploration into algebraic manipulations and solving equations, you can refer to resources like ** Khan Academy.**