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To check this, verify the following is true: \(\begin4x &= 8\ 4(2) &= 8 \ 8 &= 8\end\) This is a true statement, so our answer is correct.Let’s try a couple more examples before moving on to more complex equations.In these equations, we will need to undo two operations in order to isolate the variable.
Solve: \(3x 2=4x-1\) Since both sides are simplified (there are no parentheses we need to figure out and no like terms to combine), the next step is to get all of the x’s on one side of the equation and all the numbers on the other side.
The same rule applies – whatever you do to one side of the equation, you must do to the other side as well!
It is possible to either move the \(3x\) or the \(4x\). Since it is positive, you would do this by subtracting it from both sides: \(\begin3x 2 &=4x-1\\ 3x 2\color &=4x-1\color\\ -x 2 & =-1\end\) Now the equation looks like those that were worked before.
The next step is to subtract 2 from both sides: \(\begin-x 2\color &= -1\color\\-x=-3\end\) Finally, since \(-x= -1x\) (this is always true), divide both sides by \(-1\): \(\begin\dfrac &=\dfrac\\ x&=3\end\) You should take a moment and verify that the following is a true statement: \(3(3) 2 = 4(3) – 1\) In the next example, we will need to use the distributive property before solving.
To solve linear equations, there is one main goal: isolate the variable. Therefore, to isolate \(x\), you must divide that side by 4.
In this lesson, we will look at how this is done through several examples. So we need to figure out how to isolate the variable. When doing this, you must remember one important rule: whatever you do to one side of the equation, you must do to the other side. \(\begin4x &= 8 \ \dfrac &= \dfrac\end\) Simplifying: \(x = \boxed\) That’s it, one step and we are done.In each case, the steps will be to first simplify both sides, then use what we have been doing to isolate the variable.We will first take an in depth look at an example to see how this all works.\(\begin5x 5\color &=-3\color\ 5x &=-8\ \dfrac&=\dfrac\ x &= \dfrac \ &=\boxed\end\) This was a tough one, so remember to check your answer and make sure no mistake was made.To do that, you will be making sure that the following is a true statement: \(3\left(-\dfrac 2\right)-1=\left(-\dfrac\right)-3\left(-\dfrac 1\right)\) (Note: it does work – but you have to be really careful about parentheses!Solve: \(5w 2 = 9\) As above, there are two operations: \(w\) is being multiplied by 5 and then has 2 added to it.We will undo these by first subtracting 2 from both sides and then dividing by 5.\(\begin 3(x 2)-1 &=x-3(x 1)\ 3x 6-1&=x-3x-3 \ 3x 5&=-2x-3\end\) Now we can add 2x to both sides.(Remember you will get the same answer if you instead subtracted 3x from both sides) \(\begin 3x 5\color &=-2x-3\color\ 5x 5& =-3\end\) From here, we can solve as we did with other two-step equations.Solve: \(3x=12\) Since \(x\) is being multiplied by 3, the plan is to divide by 3 on both sides: \(\begin3x &=12\ \dfrac &=\dfrac\ x&= \boxed\end\) To check our answer, we will let \(x = 4\) and substitute it back into the equation: \(\begin3x &= 12\3(4) &= 12 \ 12 &= 12\end\) Just as before, since this is a true statement, we know our answer is correct.In the next example, instead of the variable being multiplied by a value, a value is being subtracted from the variable.