SAT Math: Systems with No Solution or Infinitely Many Solutions

Two lines can meet once, never, or be the same line. For a system written as $a x + b y = c$ , compare the ratios.

Different slopes: one solution.
Same slope, different lines: no solution.
Same line: infinitely many solutions.

Find $k$ so the system has no solution: $2 x + 3 y = 6$ and $4 x + k y = 10$

The second $x$ coefficient is 2 times the first, so match $y$ : $k = 6$ .

With $k = 6$ , doubling the first equation gives $4 x + 6 y = 12$ , not 10. The lines are parallel, so $k = 6$ .

Find $k$ so the system has infinitely many solutions: $3 x - 2 y = 5$ and $k x - 6 y = 15$

$- 6$ is 3 times $- 2$ , and 15 is 3 times 5.

So every term must be tripled: $k = 9$ .

Common mistake

Stopping once the $x$ and $y$ coefficients match.

Matching coefficients only means the slopes are equal. Now check the constants: if they also match, the lines are the same. If they don't, there's no solution.

Try it yourself

$x + 4 y = 7$ and $2 x + k y = 20$ have no solution. Find $k$ .
Show answer
$k = 8$
$5 x - y = 3$ and $10 x - 2 y = c$ have infinitely many solutions. Find $c$ .
Show answer
$c = 6$
Solve $2 x + y = 4$ and $x - y = 1$ .
Show answer
$x = \frac{5}{3}, y = \frac{2}{3}$

Find k so the system has no solution: 2x+3y=6 and 4x+ky=10

Find k so the system has infinitely many solutions: 3x−2y=5 and kx−6y=15

Common mistake

Try it yourself

Find $k$ so the system has no solution: $2 x + 3 y = 6$ and $4 x + k y = 10$

Find $k$ so the system has infinitely many solutions: $3 x - 2 y = 5$ and $k x - 6 y = 15$