y = mx + b. Solving for unknowns, graphing lines.
Self-serve tutorial - low prerequisites, straightforward concepts.
A linear equation is the math version of “steady change.” If you add the same amount of x, y changes by the same amount every time—no surprises, no curves.
A linear equation relates x and y by a constant rate (slope) plus a starting value (intercept): y = m x + b. You can (1) solve for an unknown by isolating it with inverse operations, and (2) graph the equation as a straight line using b and m or by finding two points.
Many real situations have constant-rate behavior:
Examples:
In all these, the relationship between input x and output y is predictable in one specific way:
That “constant multiplier plus a starting value” is exactly what linear equations capture.
A linear equation (in two variables) is commonly written in slope–intercept form:
y = m x + b
Where:
You already know slope as “rise over run.” This lesson adds two key skills:
1) Algebraic isolation: solve for an unknown by undoing operations.
2) Graphing: turn y = m x + b into a line you can draw and interpret.
The word “linear” is linked to lines in the plane, but it also means the formula has variables only to the first power (no squares, roots, products like x·y, etc.).
Linear (in x):
Not linear (in x):
Think of y = m x + b as a two-step machine:
1) Multiply x by m (scale the input by a constant rate).
2) Add b (shift everything up or down).
If you plug in x = 0:
y = m·0 + b = b
So b is literally where the line crosses the y-axis.
If you increase x by 1, y increases by m:
Difference:
(m x + m + b) − (m x + b) = m
So slope m is “how much y changes per 1 unit of x.”
| Term | Symbol | Meaning | How to find it |
|---|---|---|---|
| Slope | m | rate of change (rise/run) | from two points: m = (y₂ − y₁)/(x₂ − x₁) |
| y-intercept | b | y when x = 0 | plug in x = 0, or read from form y = m x + b |
| solution | (x, y) | a point that makes the equation true | any point on the line |
A solution to y = m x + b is not one number—it’s a whole set of points (x, y). That’s why the graph is a line: infinitely many solutions.
In algebra, “solve” usually means: find the value of the unknown that makes the statement true.
You do this by using inverse operations to undo whatever is happening to the unknown.
The key idea:
Common operations and their inverses:
| Operation on x | Example | Inverse operation |
|---|---|---|
| +a | x + 5 | subtract a (−5) |
| −a | x − 5 | add a (+5) |
| ×a | 3x | divide by a (/3) |
| ÷a | x/4 | multiply by a (×4) |
A typical pattern: undo addition/subtraction first, then multiplication/division.
Sometimes you want x in terms of y (for example, given an output y, what input x produced it?).
Start:
y = m x + b
Step 1: remove b (subtract b from both sides):
y − b = m x
Step 2: remove m (divide both sides by m, assuming m ≠ 0):
x = (y − b) / m
So the inverse relationship is:
x = (y − b)/m
That’s a perfect example of algebraic isolation.
Even if you don’t see y = m x + b, many linear equations reduce to a similar idea.
Example form:
a x + c = d
Solve by undoing the +c, then undoing ×a:
ax + c = d
ax = d − c
x = (d − c)/a
Sometimes the unknown is inside parentheses:
2(x − 3) + 5 = 17
You can:
1) Distribute first, then isolate, or
2) Isolate the parentheses group first.
Both work if you stay consistent.
If m = 0, then y = m x + b becomes:
y = 0·x + b = b
That’s a horizontal line. Solving for x from y = b is different:
This is your first glimpse of how solving can yield:
After solving, substitute back into the original equation.
If both sides match, you’re done.
This is especially useful when negatives or fractions appear.
A formula can feel abstract, but a graph makes it concrete:
For linear equations, graphing is especially friendly because the graph is always a straight line.
Given:
y = m x + b
1) Plot the y-intercept (0, b).
2) Use slope m = rise/run to get a second point.
3) Draw the line through the points.
Example slope interpretations:
Because a line is determined by two points, once you have two accurate points, you can draw it.
If slope/intercept form is messy, pick x values and compute y.
Steps:
1) Choose convenient x values (often 0, 1, 2, or symmetric values like −1, 0, 1).
2) Compute y for each.
3) Plot those points and draw the line.
This method is slower but very reliable.
Sometimes it’s easy to find where the line crosses axes:
If you can find both intercepts, you get two points immediately.
Consider y = m x + b:
Changing b (holding m fixed):
Changing m (holding b fixed):
A simple comparison table:
| m value | Behavior as x increases | Visual |
|---|---|---|
| m > 0 | y increases | line goes up to the right |
| m < 0 | y decreases | line goes down to the right |
| m = 0 | y stays constant | horizontal line |
If slope is m = Δy/Δx, then:
That’s the constant-rate idea in a single line.
You might wonder about vectors like v or norms like ‖v‖. Those matter more when measuring distances or doing projections. For linear equations at this level, you mainly need coordinates (x, y) and slope.
Still, keep in mind: a point (x, y) can be seen as a vector p = ⟨x, y⟩, and a line is a set of such points. Later nodes (like systems of equations) will lean harder on that viewpoint.
Linear equations are the first serious “modeling language” in math:
In calculus, linear thinking matters because:
A common workflow:
1) Identify what x represents (input).
2) Identify what y represents (output).
3) Find the starting value (when x = 0) → that’s b.
4) Find the rate “per 1 unit of x” → that’s m.
Example pattern:
Units matter:
Often you’re given two data points (x₁, y₁) and (x₂, y₂) that lie on a line.
Step 1: compute slope:
m = (y₂ − y₁) / (x₂ − x₁)
Step 2: plug into y = m x + b using one point to solve for b.
Using (x₁, y₁):
y₁ = m x₁ + b
b = y₁ − m x₁
Now you have the line.
A system is just multiple linear equations at once, like:
y = 2x + 1
y = −x + 7
Graphically, solving the system means: find the point where the lines intersect (the (x, y) that satisfies both).
Algebraically, you set them equal or eliminate a variable.
Everything in systems depends on you being fluent with:
So this node is your foundation: once one line makes sense, two lines become a solvable “meeting point” problem.
Two lines can:
1) Intersect once → one solution
2) Be parallel (same m, different b) → no solutions
3) Be the same line (same m and b) → infinitely many solutions
You saw a preview of this earlier with m = 0 and inconsistent equations. Systems formalize and generalize that idea.
We’re told the relationship is y = 3x − 5 and the output y equals 10. Find the input x that produces that output.
Start with the equation and substitute y = 10:
10 = 3x − 5
Add 5 to both sides to undo “−5”:
10 + 5 = 3x
15 = 3x
Divide both sides by 3 to undo “×3”:
15/3 = x
x = 5
Check by substituting x = 5 back into y = 3x − 5:
y = 3(5) − 5 = 15 − 5 = 10 ✓
Insight: Isolation is just reversing the operations applied to x: subtract/add first, then multiply/divide. Checking catches small sign errors.
We want to draw the line described by y = −(1/2)x + 4.
Identify parameters:
m = −1/2, b = 4
Plot the y-intercept:
When x = 0, y = 4 → point (0, 4)
Use slope m = −1/2 = rise/run:
Rise = −1 (go down 1)
Run = 2 (go right 2)
From (0, 4), move right 2 and down 1 to get a second point:
(0 + 2, 4 − 1) = (2, 3)
Optional: repeat to get another point for accuracy:
From (2, 3) move right 2 and down 1 → (4, 2)
Draw a straight line through the plotted points (0, 4), (2, 3), (4, 2)
Insight: A negative slope means the line decreases as x increases. Using rise/run keeps you from guessing the steepness.
We’re given two points and assume they lie on a linear relationship. Find y = m x + b.
Compute the slope:
m = (y₂ − y₁)/(x₂ − x₁)
Take (x₁, y₁) = (2, 1) and (x₂, y₂) = (6, 9):
m = (9 − 1)/(6 − 2) = 8/4 = 2
Use y = m x + b with one point, say (2, 1):
1 = 2(2) + b
Solve for b:
1 = 4 + b
1 − 4 = b
b = −3
Write the equation:
y = 2x − 3
Check with the other point (6, 9):
2(6) − 3 = 12 − 3 = 9 ✓
Insight: Two points determine one line. First find m from the “change ratio,” then solve for b using a known point.
A linear equation models constant-rate change: every +1 in x changes y by the same amount m.
Slope–intercept form is y = m x + b, where m is slope and b is the y-intercept (y when x = 0).
To solve for an unknown, isolate it using inverse operations while keeping both sides balanced.
From y = m x + b, solving for x gives x = (y − b)/m (when m ≠ 0).
Graphing from y = m x + b is efficient: plot (0, b), then use rise/run from m to get another point.
A line is the set of all points (x, y) that satisfy the equation; there are infinitely many solutions.
Comparing lines previews systems: intersect once (one solution), parallel (no solution), same line (infinitely many).
Sign errors when moving terms (e.g., treating “−5” as “+5” when isolating).
Misreading slope fractions: m = 1/3 means right 3, up 1 (not right 1, up 3).
Forgetting to apply an operation to both sides of the equation during isolation.
Confusing b with the x-intercept: b is always the y-value at x = 0.
Solve for x: 4x + 7 = 31.
Hint: Undo +7 first, then undo ×4.
4x + 7 = 31
4x = 31 − 7 = 24
x = 24/4 = 6
Graph y = (3/2)x − 1 by listing three points, including the y-intercept.
Hint: Start with x = 0 for the intercept. Then pick x values that avoid fractions (like x = 2 and x = 4).
x = 0 → y = (3/2)·0 − 1 = −1 → (0, −1)
x = 2 → y = (3/2)·2 − 1 = 3 − 1 = 2 → (2, 2)
x = 4 → y = (3/2)·4 − 1 = 6 − 1 = 5 → (4, 5)
Plot (0, −1), (2, 2), (4, 5) and draw the line.
Find the equation y = m x + b of the line that passes through (−1, 4) and (3, −2).
Hint: Compute m = (y₂ − y₁)/(x₂ − x₁). Then use b = y − m x with either point.
Let (x₁, y₁) = (−1, 4), (x₂, y₂) = (3, −2)
m = (−2 − 4)/(3 − (−1)) = (−6)/4 = −3/2
Use y = m x + b with (−1, 4):
4 = (−3/2)(−1) + b = 3/2 + b
b = 4 − 3/2 = 8/2 − 3/2 = 5/2
So y = (−3/2)x + 5/2
Next: Systems of Linear Equations
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