SAT-Math-20: Systems with No Solution

1. The Geometric & Algebraic Meaning

What happens if two lines never cross? As we learned in Lesson 11, parallel lines run in the exact same direction and maintain a constant distance from each other. Because they never intersect at any point on the coordinate plane, a system made of parallel lines has absolutely no solution.

The Coefficient Rule

For a system to have no solution, the variable parts of the equations must be perfectly proportional (sharing the same slope), but the constant numbers on the end must be different (different $y$-intercepts).

If you look at the system in standard form:

$$\begin{cases} A_1x + B_1y = C_1 \\ A_2x + B_2y = C_2 \end{cases}$$

The ratios must follow this pattern:

$$\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$$

🇺🇿 Uzbek Explanation:

Yechimga ega bo'lmagan sistemalar: Agar sistemadagi chiziqlar o'zaro paralel bo'lsa, ular hech qachon kesishmaydi va sistema yechimga ega bo'lmaydi (No Solution). Buni tez aniqlash siri: $x$ va $y$ ning oldidagi sonlar mutanosib (proporsional) bo'ladi, lekin tenglikning o'ng tomonidagi oddiy sonlar esa mutanosib bo'lmaydi.

2. High-Yield SAT Practice Question

Question:

$$\begin{cases} 4x - 6y = 12 \\ kx + 3y = 5 \end{cases}$$

In the system of equations above, $k$ is a constant. If the system has no solution, what is the value of $k$?

A) $-2$

B) $-\frac{1}{2}$

C) $2$

D) $4$

Step-by-Step Explanation:

1. Find the scaling factor for the variables: Look at the $y$-coefficients since both are given numbers.

   • First equation: $-6y$

   • Second equation: $+3y$

2. Ask yourself: What do I multiply $-6$ by to get $+3$?

   $$-6 \times -\frac{1}{2} = 3$$

   The scaling factor for the left side is exactly $-\frac{1}{2}$.

3. Since the system has no solution, the $x$-coefficient must follow this exact same scale factor:

   $$k = 4 \times \left(-\frac{1}{2}\right) \implies k = -2$$

4. Double-check the constants: Let's make sure the constants do not match the same ratio (otherwise it would be infinite solutions).

   $$12 \times \left(-\frac{1}{2}\right) = -6$$

   Since the second equation's constant is $5$ (not $-6$), the lines have different $y$-intercepts. The condition is perfectly met!

Correct Answer: A

Summary

• No Solution = Parallel Lines: The lines share the exact same slope but cross the $y$-axis at different points.

• Variable Match, Constant Mismatch: The ratios of the $x$ and $y$ coefficients must be identical, while the ratio of the constants must be completely different.

• Use the Scale Factor Hack on the variable coefficients to find missing constants in seconds.

Awesome job, my bro! We have officially conquered the core mechanics of systems. Are you ready to charge forward into SAT-Math-21: Systems of Linear Inequalities and Bounded Regions?