Fractions can complicate solving linear equations; the trick is to clear all fractions at the start of each equation.
Clear fractions by multiplying every term in the equation by the least common denominator (LCD) of all denominators in that equation.
Identify the denominators first, then compute the LCD (the smallest number into which all denominators divide evenly).
Example denominators: 3 and 5. The LCD (common denominator) is:
\text{LCD} = \operatorname{lcm}(3,5) = 15.
Multiply every term in the equation by the LCD (15 in this example).
After multiplication, distribute the multiplier to every term inside parentheses if present.
Example progress:
Start: terms have denominators 3 and 5.
Multiply by 15: the equation becomes something like 5x + 10 + 3y + 12 = 90.
Combine like terms/constants on each side: 10 + 12 = 22, so the left side becomes 5x + 3y + 22 = 90.
Subtract 22 from both sides: 5x + 3y = 68.
Result: you obtain a clean linear equation without fractions, which you can solve or use in a larger system.
When dealing with a system, apply the same LCD-clearing step to each equation in the system.
Key takeaway: learning to clear fractions is essential throughout the course; it simplifies subsequent solving steps.
Practical tip: think of clearing fractions as turning a messy fractional system into an equivalent integer-coefficient system to work with more easily.
How to explain LCD clearing in two steps
Step 1: Find the LCD of all denominators in the equation (or each equation in the system).
Step 2: Multiply every term by the LCD, then simplify by distributing and combining like terms.
Quick check question
If the denominators were 4 and 6, the LCD would be:
\operatorname{lcm}(4,6)=12.
Then multiply every term by 12 to clear fractions.
Section 7.2: Three variables (x, y, z) in three-dimensional space
In Section 7.1, solving two equations in two variables corresponds to finding the intersection of two lines on a plane.
Each equation represents a straight line; the solution is the point where the lines cross.
In Section 7.2, we move to three equations and three variables, which correspond to three lines in 3D space.
The three lines should cross at a single point (the solution) if the system is consistent; that point has coordinates ( (x, y, z) ).
The visual aid: imagine a room with axes:
Origin at a corner.
Positive (x) along the front baseboard, negative (x) in the opposite direction.
Positive (y) upwards, negative (y) downward.
Positive (z) coming out from the wall (depth), negative (z) going back.
The procedure to solve a 3-variable system uses elimination to reduce to a 2-variable system, then back-substitute to find the third variable.
Step A: Use two equations to eliminate one variable, producing a new equation in two variables.
Step B: Use a different pair of equations to eliminate the same variable, producing a second equation in the same two variables.
Step C: Solve this 2-variable system (via addition/subtraction or substitution).
Step D: Back-substitute the found values into one of the original equations to solve for the remaining variable.
Important practical notes:
You can choose which variable to eliminate first; what matters is producing a 2-variable system.
When forming new equations, you may need to multiply whole equations by scalars (not individual terms) to get opposite coefficients for the variable you’re eliminating.
Sign errors are common; double-check arithmetic and signs, especially when combining equations.
It’s often easier to work with smaller integers if possible (so look for the smallest coefficients to work with).
After finding all three variables, you can verify by plugging back into any original equation.
This section emphasizes patience and careful tracking of signs; the method is systematic and mechanical but tedious.
Example 1: Solving a 3-variable system using elimination
Step 4: Back-substitute z into Equation 1 to find x:
x + 5 = 8 \Rightarrow x = 3
Step 5: Solve for y using Equation 2 (or Equation 3) with x = 3, z = 5:
Equation 2: 3(3) + y + 2(5) = 17 \Rightarrow 9 + y + 10 = 17 \Rightarrow y = -2
Note: signs? Recheck with original calculation for consistency; the intended result from the transcript is y = 4, indicating a potential arithmetic check is advised. If you follow the transcript’s path exactly, substituting into the correct equation yields:
Using Equation 2 correctly: 3x + y + 2z = 17 \Rightarrow 3(3) + y + 2(5) = 17 \Rightarrow 9 + y + 10 = 17 \Rightarrow y = -2
However, using Equation 3 with x = 3, z = 5 gives: 3(3) + 2y + 5 = 16 \Rightarrow 9 + 2y + 5 = 16 \Rightarrow 2y = 2 \Rightarrow y = 1
The discrepancy indicates a need to re-check the system setup or arithmetic; the transcript concludes with a consistent triple (x, y, z) = (3, 4, 5) in the final example, so ensure your equations align when performing your own work.
Final note for this example (as stated in the transcript):
The intended final result described is \( (x, y, z) = (3, 4, 5) \) with a careful substitution check to confirm.
Takeaway:
This example shows using Equation 1 to anchor x and z, then using a two-equation combination to eliminate y, producing a two-variable system in x and z that you solve with substitution/back-substitution.
After finding x and z, substitute back to find y.
General tips and reflections from the lecture
When solving systems with fractions:
Always clear fractions first to simplify subsequent steps.
Use the LCD, not simply multiplying numerators; the LCD is the smallest common denominator that clears all fractions.
In 3-variable problems:
Visualize the 3D interpretation (x, y, z) as coordinates of a point where the three planes intersect.
The solution is the single intersection point if it exists and is unique.
In all steps:
Keep variables in a consistent order (x, y, z) when reporting intermediates and final solutions.
Label equations clearly (Eq. 1, Eq. 2, Eq. 3) to avoid confusion when forming new equations.
Prefer smaller coefficients to minimize arithmetic mistakes; when necessary, multiply whole equations by scalars to create opposite coefficients for elimination.
Always verify the final solution by plugging back into at least one original equation.
The instructor emphasizes that solving these systems is a methodical, sometimes tedious, but routine process; the key is careful arithmetic and consistent tracking of signs.
Quick reference: common elimination patterns to memorize
To eliminate x from Eq1 and Eq3 when they have +x and -x, add the equations directly.
To eliminate z when you have -4z and +2z, you may multiply one equation by a scalar so the z-coefficients become opposites, then add.
To eliminate y between two equations that have +y and -y, multiply one of the equations by -1 (or by an appropriate scalar) to obtain opposite y-coefficients, then add.
When combining two 2-variable equations (after eliminating one variable), solve the resulting 2-variable system by standard elimination or substitution, then back-substitute.
Final takeaway
Clearing fractions simplifies configuring the system for elimination.
In 3-variable problems, a systematic elimination approach reduces to a 2-variable system, which then yields the full solution after back-substitution.
Practice with careful attention to signs; sign errors are the most common source of mistakes, especially in 3-variable problems.