Here, you will learn how to solve literal equations with some carefully chosen examples. What is a literal equation? It is an equation that has many variables and we need to solve for one of the variables.
For example, y = mx + b is a literal equation. See below how we solve for x as we solve for x in 8 = 2x + 4
y = mx + b |
8 = 2x + 4 |
We can see that the process is similar whether we are solving a literal equation or not.
All we need to do is to isolate things. For (8 - 4) / 2 = x though, we can take a step further by doing the math since we are dealing with numbers.
Therefore, x = (8 - 4) / 2 = 4 / 2 = 2
Hopefully, this example was clear enough to put you on the right track.
1)
Solve 2a + b = d for b
2a + b = d
We need to isolate b, therefore, get rid of 2a by subtracting 2a from both sides of the equation.
2a - 2a + b = d - 2a
0 + b = d - 2a
b = d - 2a
2)
Solve V = lwh for w.
We need to get rid of lh. Rewrite the equation.
V = lhw
Get rid of lh by dividing both sides of the equation by lh
V / lh = (lh / lh)w
V / lh = 1w
V / lh = w
3)
Subtract a from both sides
3x - a = a - a + b + c
3x - a = 0 + b + c
3x - a = b + c
Subtract b from both sides
3x - a - b = b - b + c
3x - a - b = 0 + c
3x - a - b = c
4)
Solve 2(x + y) = z for y
First isolate x + y by getting rid of 2. To get rid of 2, divide both sides by 2.
2 / 2 (x + y) = z / 2
1(x + y) = z / 2
x + y = z / 2
Subtract x from both sides of the equations
x - x + y = (z / 2) - x
0 + y = (z / 2) - x
y = (z / 2) - x
Sep 01, 18 04:07 PM
These heart of algebra questions will help you prepare to take the math portion of the SAT
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Sep 01, 18 04:07 PM
These heart of algebra questions will help you prepare to take the math portion of the SAT
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