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My question is - Find the values of $a,b$ if the equation $a(2x+3) + 3bx=12x+5$ has infinitely many solutions.

Please can anyone guide me to solve these types of questions? I would be really thankful.

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3 Answers

up vote 1 down vote accepted

(This essentially duplicates Brian M. Scott's answer, but with very informal language. That is, his answer is more technically correct/complete and more precise; this answer is more along the lines of how I'd talk a student through the problem.)

Let's think about what happens when you try to solve this kind of equation. First, for simplicity, let's expand the product $a(2x+3)$: $$2ax+3a+3bx=12x+5$$ Since the problem says to find the values of $a$ and $b$, it's reasonable to think that $a$ and $b$ should have some specific values. If we had values for $a$ and $b$, then $x$ would be the only variable, so when the problem talks about "infinitely many solutions" to the equation, it's talking about solving for $x$.

When we solve an equation like this one (a linear equation) for a variable—in this case, $x$—we usually do so by getting all the $x$ stuff onto one side of the equation and all the non-$x$ stuff onto the other side of the equation. Here's what I mean by $x$ stuff and non-$x$ stuff: $$\underset{x\text{ stuff}}{\underbrace{2ax}}+\underset{\text{non-}x\text{ stuff}}{\underbrace{3a}}+\underset{x\text{ stuff}}{\underbrace{3bx}}=\underset{x\text{ stuff}}{\underbrace{12x}}+\underset{\text{non-}x\text{ stuff}}{\underbrace{5}}$$ Let's subtract $12x$ from both sides of the equation to "move" the $12x$ from the right side to the left side; let's also subtract $3a$ to "move" the $3a$ from the left side to the right side: $$2ax+3bx-12x=5-3a$$ If $a$ and $b$ were numbers, we'd be able to put all that $x$ stuff together more easily; if we factor out the $x$ on the left side, it may look a little clearer: $$(2a+3b-12)x=5-3a$$ Imagine that you were solving this equation for $x$ with numbers instead of $a$ and $b$. At this point, if you had some non-zero number times $x$ equals any number, you'd be able to solve for a single value of $x$ by dividing both sides by that non-zero number.

So, since we want there to be infinitely many solutions, the number we're multiplying by $x$ has to be zero. $$2a+3b-12=0$$ This means that the whole left side of the equation will be zero: $$0\cdot x=0=5-3a$$ This equation is logically equivalent to the original equation (under the condition that the original equation does not have just a single solution). That is, if $0=5-3a$ is true, then the original equation is true for all $x$; if $0-5-3a$ is false, then the original equation is false for all $x$. Since we want infinitely many solutions, we want $0=5-3a$ to be true, so solve $$\left\{\begin{align}2a+3b-12&=0\\0&=5-3a\end{align}\right.$$ for $a$ and $b$.

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The equation $a(2x+3)+3bx=12x+5$ can be simplified to $2ax+3a+3bx=12x+5$, then to $(2a+3b)x+3a=12x+5$, and finally to $$(2a+3b-12)x=5-3a\;.\tag{1}$$

If $2a+3b-12\ne 0$, $(1)$ has the unique solution $$x=\frac{5-3a}{2a+3b-12}\;,$$ so we’re interested only in the cases in which $2a+3b-12=0$. If $2a+3b-12=0$, but $5-3a\ne 0$, then $(1)$ has no solution: it’s inconsistent, because no matter what you substitute for $x$, you get $0\ne 0$. On the other hand, if $2a+3b-12=0$ and $5-3a=0$, then $(1)$ is true no matter value you substitute for $x$: it just says that $0=0$. The answer to your question, therefore, is found by solving the simultaneous system

$$\left\{\begin{align*} &2a+3b-12=0\\ &5-3a=0\;. \end{align*}\right.$$

This is pretty easy, and I’ll leave it to you.

This method will work on all such problems.

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@ Brian M.Scott thanks this really helped me .But what do they mean when they say that the equation has infinitely many solutions? I get confused a lot when something like this comes in the question. –  meg_1997 Jun 1 '12 at 11:57
@user1396721: They mean that there are infinitely many different numbers that can be substituted for $x$ to make the equation true. Consider, for instance, the silly equation $2x=x+x$: no matter what you substitute for $x$, this will be true, because it’s equivalent to $0=0$. In your problem, once you find $a$ and $b$ satisfying that system up there, your equation will also collapse to $0=0$: it will be true no matter what number you substitute for $x$. Each of the infinitely many real numbers that you could substitute for $x$ therefore is a solution, and so there are infinitely many. –  Brian M. Scott Jun 1 '12 at 12:03
@ Brian M Scott well i completely got the solution now.So whenever they say that the equation has infinitely many solutions then like in (1) apart from the variable everything must be equal to zero : –  meg_1997 Jun 1 '12 at 12:10
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You will have infinitely many solutions if the graph of the line on the left lays directly over the graph of the line on the right.

Another way to look at it is: if all the terms with $x$ add up to something nonzero, then you will be able to solve $x=$__. But if all of the terms with $x$ add up to zero, and the leftover constants say $4=5$, the system has no solutions.

Therefore you are looking for a case when the $x$ terms add up to zero, and the constant terms add up to zero... This is easy to do by subtracting everything over to the left, and then solving the resulting system of equations.

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