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Algebraic solution by cross-multiplication

Cross-multiplication is a method used to solve a pair of linear equations written in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

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Student-friendly explanation

This method uses ratios formed from the coefficients and constants to directly write the values of x and y. It is especially useful when the pair is already in standard form. Students should understand the formula and apply the signs carefully because one wrong sign changes the answer.

How to write this in exams

  1. 1

    Start with the exact idea

    Cross-multiplication is a method used to solve a pair of linear equations written in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

  2. 2

    Then show how to use it

    1. Put both equations in ax + by + c = 0 form. 2. Write a1, b1, c1 and a2, b2, c2 carefully. 3. Apply the x and y formulas with signs. 4. Simplify the fractions. 5. Verify the final values in both equations.

  3. 3

    Add one concrete example

    For 2x + 3y - 7 = 0 and 3x - 2y - 4 = 0, the cross-multiplication formula can be used to find x and y.

  4. 4

    Avoid this incomplete answer

    A common wrong answer is to swap b1 and b2 or ignore the minus signs in c1 and c2. That gives a numerically neat but wrong answer.

Definition

Cross-multiplication is a method used to solve a pair of linear equations written in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

Example

For 2x + 3y - 7 = 0 and 3x - 2y - 4 = 0, the cross-multiplication formula can be used to find x and y.

Rule to remember

If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then x = (b1c2 - b2c1)/(a1b2 - a2b1) and y = (c1a2 - c2a1)/(a1b2 - a2b1).

Memory hook

Standard form first, then cross carefully.

Examples and method

Worked example

For 2x + 3y - 7 = 0 and 3x - 2y - 4 = 0, the denominator is 2(-2) - 3(3) = -13. Then x = [3(-4) - (-2)(-7)] / -13 = 2, and y = [(-7)(3) - (-4)(2)] / -13 = 1.

Method to apply

1. Put both equations in ax + by + c = 0 form. 2. Write a1, b1, c1 and a2, b2, c2 carefully. 3. Apply the x and y formulas with signs. 4. Simplify the fractions. 5. Verify the final values in both equations.

Diagram support

No graph is needed for the method, but students should arrange the coefficients in a neat table before substitution.

How CBSE asks it

The exam may ask you to solve the pair by cross-multiplication or to identify the standard form needed for this method.

Avoid common mistakes

Common confusion

Students often place the coefficients in the wrong positions in the formula or miss the minus signs in the numerators.

Common wrong answer

A common wrong answer is to swap b1 and b2 or ignore the minus signs in c1 and c2. That gives a numerically neat but wrong answer.

Exam tip

Write the coefficients neatly first and keep the standard form consistent before applying the formula.

Quick check

Why is cross-multiplication used only after writing equations in standard form?

It is used after standard form because the formula depends on the coefficients a, b, and c being identified clearly. If the equations are not arranged properly, the sign and position errors become common.

Answer writing and exam use

1-mark answer

Cross-multiplication is a method used to solve a pair of linear equations written in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

2-mark answer

Cross-multiplication is a method used to solve a pair of linear equations written in the form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then x = (b1c2 - b2c1)/(a1b2 - a2b1) and y = (c1a2 - c2a1)/(a1b2 - a2b1). For 2x + 3y - 7 = 0 and 3x - 2y - 4 = 0, the cross-multiplication formula can be used to find x and y.

3-mark answer

This method uses ratios formed from the coefficients and constants to directly write the values of x and y. It is especially useful when the pair is already in standard form. Students should understand the formula and apply the signs carefully because one wrong sign changes the answer. If a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then x = (b1c2 - b2c1)/(a1b2 - a2b1) and y = (c1a2 - c2a1)/(a1b2 - a2b1). For 2x + 3y - 7 = 0 and 3x - 2y - 4 = 0, the denominator is 2(-2) - 3(3) = -13. Then x = [3(-4) - (-2)(-7)] / -13 = 2, and y = [(-7)(3) - (-4)(2)] / -13 = 1. The exam may ask you to solve the pair by cross-multiplication or to identify the standard form needed for this method. A common wrong answer is to swap b1 and b2 or ignore the minus signs in c1 and c2. That gives a numerically neat but wrong answer.
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