Solve 4a + 6b = 10: A Clear, Step-by-Step Guide

Solve 4a + 6b = 10: A Clear, Step-by-Step Guide

The equation 4a + 6b = 10 is a linear equation in two variables. It has infinitely many solutions in the real numbers, each forming a line when graphed. This guide explains how to simplify, solve for a specific variable, find integer solutions, interpret the graph, and avoid common mistakes. It assumes real-number solutions unless otherwise stated.

Standard Form and Key Properties

Standard form of a linear equation is Ax + By = C, where A, B, and C are real numbers and A and B are not both zero. For 4a + 6b = 10:

• Coefficient of a: 4
• Coefficient of b: 6
• Constant term: 10

Important properties:

• Linearity: The graph is a straight line in the ab-plane.
• Infinite solutions: Any point (a, b) lying on the line satisfies the equation.
• Equivalent forms: Dividing all terms by 2 yields the simpler form 2a + 3b = 5, which describes the same line.

Equivalent Simplified Equation

Divide both sides by 2:

4a/2 + 6b/2 = 10/2 → 2a + 3b = 5

The simplified equation is easier to work with and reduces the chance of arithmetic errors.

Solve for a Given b (Isolate a)

To express a in terms of b, isolate a on one side:

1. Start with 4a + 6b = 10.
2. Subtract 6b from both sides: 4a = 10 − 6b.
3. Divide by 4: a = (10 − 6b) / 4, which simplifies to a = 2.5 − 1.5b.

Now, for any real number b, you can compute the corresponding a that satisfies the equation. For example, if b = 1, then a = 2.5 − 1.5(1) = 1, giving the solution pair (1, 1).

Solve for b Given a (Isolate b)

To express b in terms of a, isolate b on one side:

1. Start with 4a + 6b = 10.
2. Subtract 4a from both sides: 6b = 10 − 4a.
3. Divide by 6: b = (10 − 4a) / 6, which simplifies to b = (5/3) − (2/3)a.

Now, for any real number a, you can compute the corresponding b. For example, if a = 2, then b = (5/3) − (2/3)(2) = 1/3, giving the solution pair (2, 1/3).

Find Integer and Simple Rational Solutions

Integer solutions occur when both a and b are integers. From 2a + 3b = 5, note that 2a = 5 − 3b, so 5 − 3b must be even. Since 5 is odd, 3b must be odd, meaning b must be odd.

Testing small integer values of b:

• b = 1 → 2a + 3 = 5 → 2a = 2 → a = 1. Solution: (1, 1).
• b = −1 → 2a − 3 = 5 → 2a = 8 → a = 4. Solution: (4, −1).
• b = 3 → 2a + 9 = 5 → 2a = −4 → a = −2. Solution: (−2, 3).

These are integer solutions. There are infinitely many other solutions with non-integer real values.

Interpret the Graph in the ab-Plane

Graphically, 4a + 6b = 10 is a line. Using the slope-intercept form b = (5/3) − (2/3)a:

• Slope: −2/3, meaning b decreases by 2/3 for every 1-unit increase in a.
• b-intercept: (0, 5/3), the point where the line crosses the b-axis.
• a-intercept: (2.5, 0), found by setting b = 0 in 4a = 10.

Plotting these two intercepts and drawing a line through them gives all solutions. Each point on the line corresponds to one valid (a, b) pair.

Common Mistakes and Checks

• Mistake: Dividing incorrectly when simplifying. Always divide every term by the same nonzero number.
• Mistake: Sign errors when moving terms across the equals sign. Double-check subtraction and distribution.
• Check: Substitute your pair (a, b) back into 4a + 6b to verify it equals 10.
• Check: Confirm intercepts by plugging a = 0 or b = 0 into the original equation.

Summary of Key Points

When to Use This Approach

This method applies to any linear equation in two variables. Use it to:

• Express one variable in terms of the other for modeling relationships.
• Find particular solutions that meet constraints (e.g., integer or non-negative values).
• Interpret trends and trade-offs by analyzing slope and intercepts.