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(That’s why equations like these are often called “one-step” equations) Anytime you are solving linear equations, you can always check your answer by substituting it back into the equation.
\(\begin 3(x 2)-1 &=x-3(x 1)\\ 3x 6-1&=x-3x-3 \\ 3x 5&=-2x-3\end\) 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.
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: \(3x 2=4x-1\) 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!
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.