Consistency and the ratio test

The geometric story has three cases. The algebraic test that decides which case you are in is the ratio test, and it is one of the cleanest theorems in the chapter.

Definitions

A pair of linear equations is:

• Consistent if it has at least one solution.
• Inconsistent if it has no solution.
• Independent if it has a unique solution; dependent if it has infinitely many.

Theorem: the ratio test

Consider the pair $a_1 x + b_1 y + c_1 = 0,$ $a_2 x + b_2 y + c_2 = 0.$

Compute the three ratios $a_1/a_2$, $b_1/b_2$, $c_1/c_2$ (whenever defined).

Why it works. The slope of $a x + b y + c = 0$ is $-a/b$ (assuming $b \ne 0$), and the $y$-intercept is $-c/b$.

• Different slopes ($-a_1/b_1 \ne -a_2/b_2$, equivalently $a_1/a_2 \ne b_1/b_2$) $\Rightarrow$ lines cross.
• Same slope, different $y$-intercepts $\Rightarrow$ parallel and distinct.
• Same slope, same $y$-intercept $\Rightarrow$ coincident.

In ratio form, "same slope" becomes $a_1/a_2 = b_1/b_2$, and the third ratio decides whether the lines also share the intercept.

The test is mechanical. You do not have to solve the system to know how many solutions exist. This is invaluable in problems asking "for which values of $k$ does the system have no solution?" , you just set the ratios appropriately.

A useful intuition: think of the three ratios as a signature. If the first two ratios differ, the system has a unique solution and the third ratio is irrelevant. If the first two ratios agree, then the third ratio decides between "no solution" and "infinitely many".

Worked examples

Example 1. Classify the system $3 x + 2 y = 8$, $6 x + 4 y = 16$.

Rewrite with all terms on one side: $3 x + 2 y - 8 = 0$ and $6 x + 4 y - 16 = 0$.

Ratios: $3/6 = 1/2$, $2/4 = 1/2$, $-8/-16 = 1/2$. All three equal $\Rightarrow$ infinitely many solutions; lines coincide.

Example 2. Classify the system $x + 2 y = 5$, $2 x + 4 y = 11$.

Ratios: $1/2$, $2/4 = 1/2$, $-5/-11 = 5/11$. First two equal but the third differs $\Rightarrow$ no solution; lines parallel.

Example 3. Classify $2 x + 3 y = 7$, $3 x + 4 y = 11$.

$2/3$ vs $3/4$. Not equal $\Rightarrow$ unique solution; lines intersect.

Example 4. For what value of $k$ does the system $x + 2 y = 3$, $5 x + k y = 15$ have infinitely many solutions?

Need $\dfrac{1}{5} = \dfrac{2}{k} = \dfrac{3}{15}$. From $1/5 = 2/k$: $k = 10$. Check the third: $3/15 = 1/5$. ✓ So $k = 10$.

Example 5. For what value of $k$ are the lines $2 x + 3 y = 7$ and $(k - 1) x + (k + 2) y = 3 k$ inconsistent?

Need $\dfrac{2}{k - 1} = \dfrac{3}{k + 2} \ne \dfrac{7}{3k}$.

From $2/(k-1) = 3/(k+2)$: $2(k+2) = 3(k-1) \Rightarrow 2 k + 4 = 3 k - 3 \Rightarrow k = 7$.

Check the third ratio: $\dfrac{2}{7-1} = \dfrac{1}{3}$ and $\dfrac{7}{3 \cdot 7} = \dfrac{1}{3}$. They are equal, so the system would actually be consistent (infinitely many) for $k = 7$, not inconsistent.

So no value of $k$ makes the system inconsistent. (Important to verify the third ratio.)

Try it yourself

1. Classify each as unique / no / infinite solutions: (a) $x + y = 5, 2 x - y = 1$ (b) $2 x - 3 y = 5, 4 x - 6 y = 7$ (c) $3 x + 4 y = 12, 6 x + 8 y = 24$
2. Find $k$ so that $2 x + k y = 5, 3 x + 2 y = 7$ has a unique solution.
3. Find $k$ so that $k x + 3 y = k - 3, 12 x + k y = k$ has infinitely many solutions.
4. Find $k$ so that $3 x + y = 1, (2 k - 1) x + (k - 1) y = 2 k + 1$ is inconsistent.
5. For what value of $\alpha$ are the lines $\alpha x + 3 y = \alpha - 3$ and $12 x + \alpha y = \alpha$ coincident?
6. Show that the system $3 x - 5 y = 4, 9 x - 15 y = 18$ is inconsistent without solving.
7. For what value of $k$ is $kx + 2y = 5, 3x + y = 1$ consistent independent?
8. The system $2 x + 3 y = 5, p x - 6 y = 8$ has no solution. Find $p$.
9. Determine without graphing whether $3 x + 2 y = 5, 2 x - 3 y = 7$ has a unique solution.
10. For what value(s) of $k$ does $kx - y = 2, 6x - 2y = 3$ have (a) unique solution, (b) no solution, (c) infinitely many?

Pitfalls / Insight

• Always bring both equations to the form $a x + b y + c = 0$ before computing ratios. Don't mix forms.
• Check the third ratio when the first two agree , otherwise you'll mis-call "no solution" as "infinite solutions" and vice versa.
• Be careful when a coefficient is $0$: e.g., if $b_1 = 0$, the ratio $b_1/b_2$ is $0$ and the comparison may need a small fix-up.

Insight. The ratio test is a one-glance classification. Use it before solving , sometimes the problem says "no solution" and you can immediately set the ratios appropriately, saving minutes of algebra.