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Since you are undoing the operations to \(x\), you will work from the “outside in”.
It is easy to make a mistake here, so make sure that you distribute the number in front of the parentheses to all the terms inside.
Solve: \(3(x 2)-1=x-3(x 1)\) First, distribute the 3 and –3, and collect like terms.
Let’s look at one more two-step example before we jump up in difficulty again.
Make sure that you understand each step shown and work through the problem as well.
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.
\(\begin 3(x 2)-1 &=x-3(x 1)\ 3x 6-1&=x-3x-3 \ 3x 5&=-2x-3\end\) 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: \(2x-7=13\) Notice the two operations happening to \(x\): it is being multiplied by 2 and then having 7 subtracted. But, only the \(x\) is being multiplied by 2, so the first step will be to add 7 to both sides. Adding 7 to both sides: \(\begin 2x-7 &= 13\ 2x-7 \color & =13 \color\ 2x&=20\end\) Now divide both sides by 2: \(\begin 2x &=20 \ \dfrac&=\dfrac\ x&= \boxed\end\) Just like with simpler problems, you can check your answer by substituting your value of \(x\) back into the original equation.
\(\begin2x-7&=13\ 2(10) – 7 &= 13\ 13 &= 13\end\) This is true, so we have the correct answer.