S1Q2 · Parse Equation and Solve for x

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Function: solve_for_x(equation: str) -> float

Core idea: split on x to get the coefficient of x and the constant; split on = to get the right-hand side. Solve ax + b = c → x = (c - b) / a.

def solve_for_x(equation: str) -> float:
    equation = equation.replace(" ", "")
    left, c = equation.split("=")
    a_part, b_part = left.split("x")
    a = float(a_part) if a_part not in ("", "+", "-") else float(a_part + "1")
    b = float(b_part) if b_part != "" else 0.0
    return (float(c) - b) / a

Key rules: - Strip all spaces first with replace(" ", "") - Split on "x" → left part is a, right part is b - If nothing before x → a = 1; if just "-" before x → a = -1 - If nothing after x (before =) → b = 0 - Formula: x = (c - b) / a

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Problem Statement

Problem

Write a function solve_for_x(equation: str) -> float that parses a linear equation of the form ax+b=c or ax-b=c and returns the value of x as a float. Spaces may appear anywhere and must be ignored.

Examples:

Input

"2x +3= 11"

Output

4.0

Input

"-3x + 10=1"

Output

3.0

Input

"x + 2 = 5"

Output

3.0

Input

"2x=6"

Output

3.0

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Understanding the problem

The equation always has the form ax + b = c. The goal is to extract a, b, and c, then compute x = (c - b) / a.

The hint says to treat x and = as separators:

"2x+3=11"
   ↑   ↑
   x   =

Split on "=":   left = "2x+3",   right = "11"  → c = 11
Split on "x":   a_part = "2",    b_part = "+3"  → a = 2, b = 3

x = (11 - 3) / 2 = 4.0 

Edge cases to handle

Situation	Example	What it means
No coefficient before x	"x+2=5"	a = 1
Negative coefficient before x	"-x+2=5"	a = -1
No constant term	"2x=6"	b = 0
Negative constant	"5x-2=13"	b = -2

The a_part special cases

After splitting on "x", the part before x can be:

• "2" → a = 2
• "" → a = 1 (no coefficient means 1)
• "-" → a = -1 (just a minus sign means -1)
• "+" → a = 1 (just a plus sign means +1, rare but possible)

The cleanest fix: if a_part ends in a sign with no digits, append "1".

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Tracing all examples

After removing spaces:

Equation	Split on =	Split on x	a	b	c	x = (c-b)/a
"2x+3=11"	"2x+3", "11"	"2", "+3"	2	3	11	(11-3)/2 = 4.0
"5x-2=13"	"5x-2", "13"	"5", "-2"	5	-2	13	(13-(-2))/5 = 3.0
"-3x+10=1"	"-3x+10", "1"	"-3", "+10"	-3	10	1	(1-10)/-3 = 3.0
"x+2=5"	"x+2", "5"	"", "+2"	1	2	5	(5-2)/1 = 3.0
"2x=6"	"2x", "6"	"2", ""	2	0	6	(6-0)/2 = 3.0

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Solution approaches

Explanatory (manual parsing)Pythonic (sign + 1 trick)Using regexUsing lambda for sign handling

def solve_for_x(equation: str) -> float:
    equation = equation.replace(" ", "")      # strip all spaces

    left, c = equation.split("=")             # split on =
    a_part, b_part = left.split("x")          # split on x

    # handle coefficient of x
    if a_part in ("", "+"):
        a = 1.0
    elif a_part == "-":
        a = -1.0
    else:
        a = float(a_part)

    # handle constant term
    b = float(b_part) if b_part != "" else 0.0

    return (float(c) - b) / a

Each edge case handled explicitly -most readable.

def solve_for_x(equation: str) -> float:
    equation = equation.replace(" ", "")
    left, c = equation.split("=")
    a_part, b_part = left.split("x")

    # if a_part is empty or just a sign, append "1"
    if a_part in ("", "+", "-"):
        a_part = a_part + "1"

    a = float(a_part)
    b = float(b_part) if b_part != "" else 0.0
    return (float(c) - b) / a

Appending "1" to a bare sign ("" → "1", "-" → "-1") lets float() handle it cleanly.

import re

def solve_for_x(equation: str) -> float:
    equation = equation.replace(" ", "")
    # match: optional sign+digits before x, optional sign+digits after x, = sign+digits
    match = re.match(r"([+-]?\d*)x([+-]\d+)?=([+-]?\d+)", equation)
    a_str = match.group(1)
    b_str = match.group(2)
    c_str = match.group(3)

    a = float(a_str) if a_str not in ("", "+", "-") else float((a_str or "") + "1")
    b = float(b_str) if b_str else 0.0
    return (float(c_str) - b) / a

Uses a regex pattern to capture the three parts in one pass. Overkill for this problem but useful if the equation format is more varied.

def solve_for_x(equation: str) -> float:
    equation = equation.replace(" ", "")
    left, c = equation.split("=")
    a_part, b_part = left.split("x")

    parse = lambda s, default: float(s) if s not in ("", "+", "-") \
                               else float((s or "") + "1") if s in ("", "+", "-") \
                               else default

    a = float(a_part + "1") if a_part in ("", "+", "-") else float(a_part)
    b = float(b_part) if b_part != "" else 0.0
    return (float(c) - b) / a

A lambda encapsulates the sign-edge-case logic, making parse(a_part, 1.0) and parse(b_part, 0.0) self-documenting calls.

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Key takeaways

01

Strip spaces first

equation.replace(" ", "") removes all spaces in one call before any parsing. Always sanitise input before splitting -spaces inside tokens break float().

02

Use the problem's natural separators

The hint says treat x and = as separators. split("=") then split("x") cleanly extracts all three parts without a complex parser.

03

Handle the implicit coefficient 1

When there's no digit before x, the coefficient is 1. The sign-append trick (a_part + "1") handles "", "+", and "-" uniformly.