Cross-Multiplication, Conditions for Consistency and Reducible Equations

The Cross-Multiplication Method

For the pair $$a_1 x + b_1 y + c_1 = 0$$ $$a_2 x + b_2 y + c_2 = 0$$ there is a direct formula for the solution, called the cross-multiplication method:

$$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$$

From this: $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \qquad y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$$

Key Point: Both equations must be in the form $ax + by + c = 0$ (everything on one side) before applying the formula. A unique solution exists when the denominator $a_1 b_2 - a_2 b_1 \neq 0$.

[Board Important] This method is in many State Board syllabi (removed from rationalised CBSE). It's especially fast when you only need $x$ or only $y$.

A Memory Layout for Cross-Multiplication

The formula is easier to remember with the criss-cross diagram. Write the coefficients in two rows, repeating the $b$ and $c$ columns to make the cross pattern:

$$\begin{array}{cccc} b_1 & c_1 & a_1 & b_1 \\ b_2 & c_2 & a_2 & b_2 \end{array}$$

• For $x$: cross-multiply the first block $(b_1 c_2 - b_2 c_1)$.
• For $y$: cross-multiply the middle block $(c_1 a_2 - c_2 a_1)$.
• For $1$: cross-multiply the last block $(a_1 b_2 - a_2 b_1)$.

Key Point: Each numerator is a 'cross' difference: down-product minus up-product. Keeping the columns in the order $b, c, a, b$ gives all three correctly.

[Board Important] Practise the layout a few times — once memorised, cross-multiplication is the quickest route to both $x$ and $y$ in one go.

Conditions for Consistency (Ratio Test)

The ratios of coefficients tell us, without solving, how many solutions a pair has. This ties together everything from the graphical section.

Condition	Lines	Solutions	Type
$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$	Intersecting	Unique	Consistent
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$	Coincident	Infinitely many	Consistent (dependent)
$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$	Parallel	None	Inconsistent

Key Point: Write both equations as $ax + by + c = 0$ first, then compare the three ratios. This is the fastest way to answer 'how many solutions?' or 'find $k$ for … solutions' questions.

[Board Important] 'Find the value of $k$ for which the pair has no solution / infinitely many solutions' is asked almost every year. Use the ratio conditions directly.

Equations Reducible to Linear Form

Some equations don't look linear but become linear after a clever substitution. The classic type involves $\dfrac{1}{x}$ and $\dfrac{1}{y}$.

The trick

For a pair like $$\frac{2}{x} + \frac{3}{y} = 13, \qquad \frac{5}{x} - \frac{4}{y} = -2,$$ let $u = \dfrac{1}{x}$ and $v = \dfrac{1}{y}$. The pair becomes $$2u + 3v = 13, \qquad 5u - 4v = -2,$$ which is an ordinary linear pair. Solve for $u, v$, then recover $x = \dfrac{1}{u}$, $y = \dfrac{1}{v}$.

Key Point: Substituting $u = 1/x$, $v = 1/y$ (or similar) turns a non-linear-looking pair into a linear one. Don't forget to convert back at the end.

[Board Important] Always state your substitution clearly ('Let $u = 1/x$…') and remember to find the original variables $x$ and $y$ at the end — a common place to lose marks.

Solved Examples

Example 1: Cross-multiplication

Solve by cross-multiplication: $2x + 3y - 7 = 0$ and $3x - y - 5 = 0$.

Solution:

1. Here $a_1=2, b_1=3, c_1=-7$; $a_2=3, b_2=-1, c_2=-5$.
2. $a_1 b_2 - a_2 b_1 = (2)(-1) - (3)(3) = -2 - 9 = -11$.
3. $x = \dfrac{b_1 c_2 - b_2 c_1}{-11} = \dfrac{(3)(-5) - (-1)(-7)}{-11} = \dfrac{-15 - 7}{-11} = 2$.
4. $y = \dfrac{c_1 a_2 - c_2 a_1}{-11} = \dfrac{(-7)(3) - (-5)(2)}{-11} = \dfrac{-21 + 10}{-11} = 1$.

Final Answer: $x = 2$, $y = 1$.

Takeaway: Plug into the cross-multiplication formula carefully with signs.

Example 2: Unique solution check

Does $3x + 2y = 5$ and $2x - 3y = 7$ have a unique solution?

Solution:

1. $\dfrac{a_1}{a_2} = \dfrac{3}{2}$, $\dfrac{b_1}{b_2} = \dfrac{2}{-3}$.
2. Since $\dfrac{3}{2} \neq \dfrac{2}{-3}$, the ratios differ.

Final Answer: Yes, a unique solution.

Takeaway: $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$ ⇒ unique solution.

Example 3: Find $k$ for no solution

Find $k$ so that $3x + ky = 7$ and $6x + 10y = 11$ has no solution.

Solution:

1. No solution needs $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$.
2. $\dfrac{3}{6} = \dfrac{k}{10} \Rightarrow k = 5$.
3. Check $\dfrac{c_1}{c_2}$: $\dfrac{-7}{-11} = \dfrac{7}{11} \neq \dfrac{1}{2}$. ✓ (different)

Final Answer: $k = 5$.

Takeaway: Match the first two ratios for $k$; confirm the third is different.

Example 4: Find $k$ for infinitely many solutions

Find $k$ so that $2x + 3y = 7$ and $(k-1)x + (k+2)y = 3k$ has infinitely many solutions.

Solution:

1. Need $\dfrac{2}{k-1} = \dfrac{3}{k+2} = \dfrac{7}{3k}$.
2. From the first two: $2(k+2) = 3(k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7$.
3. Check with third: $\dfrac{2}{6} = \dfrac{1}{3}$ and $\dfrac{7}{21} = \dfrac{1}{3}$. ✓

Final Answer: $k = 7$.

Takeaway: Solve from one pair of ratios, then verify the remaining one.

Example 5: Reducible — basic

Solve: $\dfrac{1}{x} + \dfrac{1}{y} = 5$ and $\dfrac{1}{x} - \dfrac{1}{y} = 1$.

Solution:

1. Let $u = \dfrac{1}{x}$, $v = \dfrac{1}{y}$: $u + v = 5$, $u - v = 1$.
2. Add: $2u = 6 \Rightarrow u = 3$; subtract: $2v = 4 \Rightarrow v = 2$.
3. So $x = \dfrac{1}{3}$, $y = \dfrac{1}{2}$.

Final Answer: $x = \dfrac{1}{3}$, $y = \dfrac{1}{2}$.

Takeaway: Substitute $u = 1/x$, $v = 1/y$, solve, then invert.

Example 6: Reducible — harder

Solve: $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = -2$.

Solution:

1. Let $u = \dfrac{1}{x}$, $v = \dfrac{1}{y}$: $2u + 3v = 13$, $5u - 4v = -2$.
2. Eliminate $u$: multiply first by 5, second by 2: $10u + 15v = 65$, $10u - 8v = -4$. Subtract: $23v = 69 \Rightarrow v = 3$.
3. $2u + 9 = 13 \Rightarrow u = 2$. So $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Final Answer: $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Takeaway: After substitution, use elimination as usual, then invert $u, v$.

Example 7: Consistency classification

Classify $5x - 4y + 8 = 0$ and $7x + 6y - 9 = 0$.

Solution:

1. $\dfrac{a_1}{a_2} = \dfrac{5}{7}$, $\dfrac{b_1}{b_2} = \dfrac{-4}{6} = -\dfrac{2}{3}$.
2. Since $\dfrac{5}{7} \neq -\dfrac{2}{3}$, the lines intersect.

Final Answer: Unique solution; consistent.

Takeaway: Different first two ratios ⇒ consistent with a unique solution.

Example 8: Find $k$ for unique solution

For what value of $k$ does $kx + 3y = k - 3$ and $12x + ky = k$ have a unique solution?

Solution:

1. Unique solution needs $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$: $\dfrac{k}{12} \neq \dfrac{3}{k}$.
2. $k^2 \neq 36 \Rightarrow k \neq \pm 6$.

Final Answer: All $k$ except $k = 6$ and $k = -6$.

Takeaway: Cross-multiply the inequality to find the excluded values.

Example 9: Reducible with constants on both terms

Solve: $\dfrac{3}{x} + \dfrac{2}{y} = 12$ and $\dfrac{2}{x} + \dfrac{3}{y} = 13$.

Solution:

1. Let $u = 1/x$, $v = 1/y$: $3u + 2v = 12$, $2u + 3v = 13$.
2. Multiply first by 3, second by 2: $9u + 6v = 36$, $4u + 6v = 26$. Subtract: $5u = 10 \Rightarrow u = 2$.
3. $3(2) + 2v = 12 \Rightarrow v = 3$. So $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Final Answer: $x = \dfrac{1}{2}$, $y = \dfrac{1}{3}$.

Takeaway: Same substitution trick works whenever only $1/x$ and $1/y$ appear.

Example 10: Cross-multiplication for a word setup

Solve by cross-multiplication: $x + y - 5 = 0$ and $2x - y - 1 = 0$.

Solution:

1. $a_1=1, b_1=1, c_1=-5$; $a_2=2, b_2=-1, c_2=-1$.
2. $a_1 b_2 - a_2 b_1 = (1)(-1) - (2)(1) = -3$.
3. $x = \dfrac{(1)(-1) - (-1)(-5)}{-3} = \dfrac{-1 - 5}{-3} = 2$.
4. $y = \dfrac{(-5)(2) - (-1)(1)}{-3} = \dfrac{-10 + 1}{-3} = 3$.

Final Answer: $x = 2$, $y = 3$.

Takeaway: Cross-multiplication gives both variables directly from the coefficients.