Two straight lines are parallel exactly when they share the same gradient. Since they rise and fall at identical rates, they never meet, however far they're extended:

The two lines must have different $y$-intercepts — otherwise they'd be the same line, not two parallel ones. Geometrically, this means one line is simply a vertical "shift" of the other, without any change in its tilt.

1. Find the gradient $m$ of the given line, rearranging into $y=mx+c$ form first if needed.
2. Use that same gradient for the new line.
3. Substitute the coordinates of the point the new line must pass through to find its own $c$.

Find the equation of the line parallel to $y=4x-1$ that passes through $(1,-3)$:

1. Gradient of the given line: $m=4$.
2. The new line also has $m=4$: $y=4x+c$.
3. Substitute $(1,-3)$: $-3 = 4(1)+c \implies c=-7$.
4. Final equation:

Rearrange both equations into $y=mx+c$ form, then compare gradients — if $m_1=m_2$, the lines are parallel; if not, they aren't.

Determine whether $2x+y=7$ and $y=2x+5$ are parallel:

1. Rearrange the first: $2x+y=7 \implies y=-2x+7 \implies m_1=-2$.
2. Read the second directly: $m_2=2$.
3. Since $-2 \neq 2$, the lines are not parallel.

Find the equation of the required parallel line first (matching gradient, then solving for $c$ using the point you know it must pass through), then substitute the coordinates of the point in question and check whether the equation still balances.

A line parallel to $y=3x+2$ passes through $(0,4)$. Does the point $(2,10)$ lie on this new line?

1. The new line has gradient $3$ and passes through $(0,4)$, so its intercept is already $c=4$: $y=3x+4$.
2. Substitute $x=2$: $y=3(2)+4=10$.
3. Since this matches the given $y$-coordinate exactly, $(2,10)$ does lie on the line.

Find the equation of the line parallel to $3x+y=9$ that passes through $(-2,5)$.

Solution

1. Rearrange the given line into $y=mx+c$ form: $3x+y=9 \implies y=-3x+9 \implies m=-3$.
2. The new line has the same gradient: $y=-3x+c$.
3. Substitute $(-2,5)$: $5 = -3(-2)+c \implies 5=6+c \implies c=-1$.
4. Final equation:

The coefficient of $x$ in an equation's original form is often not its actual gradient. Only once an equation is correctly isolated into $y=mx+c$ form does the number in front of $x$ genuinely represent the gradient — comparing raw coefficients from unrearranged equations is one of the most common ways marks are lost on this topic.