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$6 = 4 - 2x =$

Show the answer with the mechanics of working out.

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Presumably the exercise wants you to solve for $x$. But that is a guess, since quite non-standard (I am being excessively polite) notation is being used. – André Nicolas May 25 '13 at 1:45
In the way you present this there is no meaning. Sometimes we "chain" equalities together, like $(x+y)^2=(x+y)(x+y)=x^2+2xy+y^2$, however, there's nothing after the second equality sign, so as this stands there's no meaning. – user1620696 May 25 '13 at 2:29

Equality: "$=$" is a symmetric relation, as are all equivalence relations: $$a = b \quad \overbrace{\iff}^{\text{equivalent}}\quad b = a$$ That is, it doesn't matter if we replace $\;$ "$a$ is equal to $b$" $\;$ with $\;$ "$b$ is equal to $a$:" $\;$ both expressions tell us precisely the same thing.

So, the following two equations (left hand side & right hand side of the arrow) are equivalent (meaning, really, that they give the same identical information: $$6 = 4 - 2x \iff 4 - 2x = 6$$

Now, I am assuming that you need to solve for $x$. Let's just go ahead and use the form of the equation you were given:

  • $(1)$ add $+ 2x$ to each side of the equation.$$\begin{align} 6 & = 4 - 2x \\ \\ 2x + 6 & = 4 - 2x + 2x\end{align}$$ Simplify.

  • $(2)$ Subtract $6$ from each side of the equation. Simplify.

  • $(3)$ Divide each side of the equation by $2$. (Put differently, you can multiply each side of the equation by $\frac 12).$

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Thank you for your help. I was trying to find the value of X. These explanations are very good. The equals being on the left hand side rather then on the right hand side was confusing me. But now I know that it is equal as explained by amWhy – student May 25 '13 at 1:51
Let me know what you obtain, following each step above: I "showed" you the first step, now simplify. The proceed with the other steps. I'll be happy to check your progress. I'm sure you can do it! – amWhy May 25 '13 at 1:54
@amWhy: +1 for typing all of those details out! :-) – Amzoti May 25 '13 at 2:09
@amWhy Just saying, but some people who have started with algebra-precalculus may not understand what a relation is, let alone a symmetric relation. I think you should include a link which explains equivalence in classical logic as well as relations. Thanks :-) – Parth Kohli May 25 '13 at 3:01
You're right, @ΠάρτηΚοηλί: I'll include a link. – amWhy May 25 '13 at 3:06

It does not matter left or right.

We can add $2x$ on both sides to get $$6+2x=4-2x+2x$$ $$\implies 6+2x=4$$ then subtract 6 on both sides we get $$6-6+2x=4-6$$ $$\implies 2x=-2$$ then divide 2 on both sides we get $$\frac{2}{2}x=\frac{-2}{2}$$ $$\implies x=-1.$$

If we have an equation, it is "legal" to perform addition, subtraction, multiplication and division (except by 0) by equal number on both sides. Later you will also learn that it is also "legal" to take the log, power, derivative and integral and so on, on both sides.

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Given the equation: $$\boxed{6=4-2x}$$ We can subtract $-4$ from both sides: $$6-4=4-4-2x$$ Simplifying we are left with: $$2=-2x$$ Dividing by $-2$ on both sides to isolate x: $$\dfrac{2}{-2}=\dfrac{-2x}{-2}$$ Leaving us with the solution: $$\boxed{x=-1}$$

We can check we have the correct solution by plugging the $-1$ back into the original equation:




We find the answers are the same which proves we have the correct solution.

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