What’s up everyone! Let’s review what we learned in Session 8 this past week and explore how we can solve more complicated linear equations and even some intriguing word problems.
If the following problems seem too hard, check out our introductory article: Introduction to One-Variable Linear Equations.
Solving linear equations is simple. All you need to do is to get the variable by itself on one side with a constant on the other. When you need to move terms around, all you need to do is add, subtract, multiply, or divide by a constant or variable on both sides. Recall that you have to operate on both sides, not just one or the other. The equation is like a “see-saw”: if you do something to only one side, the scale is going to tip.
Let’s review with a simple example. Try it on your own for a few minutes before scrolling down!
Problem 1:
Solve for \( x \) in the following equation:
\[ 7x+18=46 \]
How should we approach this?
Remember our goal: we want the value of the variable \( x \). Thus, we want to transform this equation into \( x = ? \), where the right-hand side is some constant. We don’t know the value of that constant yet, but we can use our handy algebra skills to do so!
\( 7x + 18=46 \) … this unfortunately doesn’t look like our desired form. We have constants on both sides ( 18 and 46) and a 7 attached to the x. This is definitely not in the form \( x = ? \). Thus, let’s try cleaning the equation up by moving the constants onto one side, and the variables onto the other.
The left-hand side has an unknown value (\( 7x \) – we don’t know the numerical of this yet) and a constant (\( 18 \) ), while the right-hand side only has a constant (\(46\)). If we get the \( 18 \) to the right-hand side, we’ll have a side with only the constants and a side with only the variables.
How can we eliminate \( 18 \) from the left-hand side of the equation?
Let’s subtract \( 18 \) from both sides:
\[ (7x+18)-18 = 46-18 \]
\[ 7x + 18-18 = 46-18 \]
\[ 7x = 28 \]
Remember to do your arithmetic correctly 😉
And now look what happened! We have a one-step problem, where we have variables on one side and constants on another.
We can get rid of the coefficient of \( x \) on the left-hand side by dividing by \( 7 \) on both sides. Remember that multiplication and division are inverse operations, so when we multiply \( x \) by seven (as in the problem) and divide by seven (which we’re doing), we should end up with \( x \). Multiplication and division, in other words, “undo” each other.
\[ 7x \div 7 = 28 \div 7 \]
\[x=4 \]
We often divide both sides using the division symbol, so the step above can also look like this:
\[ \frac{7x}{7}=\frac{28}{7} \]
\[ x= 4 \]
In general, we want to move the variables and constants in such a way that we end up with an equation in the form \( ax = b \) where \( a \) and \( b \) are both numbers, and \( a \ne 0 \) (if \( a \) is zero, it’s a special case – we’ll cover that in the future). This way, we can divide by \( a \) on both sides to get the solution \( x=b \div a = \frac{b}{a} \). Again, we achieve this form by adding, subtracting, multiplying or dividing some value on both sides.
Disclaimer: there is no right sequence to solve a linear equation! You just have to make sure that you do the same operation to both sides. Remember, our goal when solving for a variable \( x \) is the same:
\[ x = ? \]
The steps above may seem unintuitive at first, but the things you do to solve a linear equation will become more natural as you do more problems. For instance, if you were to solve for \( y \) in the equation
\[ y + 4= 34, \]
it doesn’t make a whole lot of sense to multiply by two on both sides, as that would result in
\[ 2y + 8 = 68,\]
which makes us farther from our goal of \( y = ? \).
It also doesn’t make sense to subtract \( 5732 \) to both sides:
\[ y-5728=-5698 \]
You even get negative numbers… nasty.
I think you get the idea, let’s move on to a trickier example.