SAT Math Practice Test: Systems of Linear Equations

Adding the equations eliminates y: 3x = 12, so x = 4 and then y = 3. The choice 3 is the value of y, not x.

With a + c = 40 and 12a + 7c = 395, substituting c = 40 − a gives 5a + 280 = 395, so a = 23. The choice 17 is the number of child tickets.

The lines are parallel when the coefficients are proportional: 6 is 2 × 3, so k must be 2 × (−2) = −4. Since 8 is not 2 × 5, the lines are distinct and never meet. The choice 4 comes from dropping the negative sign.

Infinitely many solutions means the second equation is a multiple of the first. Since 6x + 15y is 3 times 2x + 5y, c = 3 × 7 = 21. The choice 14 uses a factor of 2 instead of 3.

Setting 3x − 5 = −x + 7 gives 4x = 12, so x = 3 and y = 3(3) − 5 = 4. The point (2, 1) lies on the first line but not on the second.

Each animal has one head, so c + g = 30, and the legs give 2c + 4g = 84. The second choice swaps the leg counts, giving chickens 4 legs and goats 2.

Adding the equations gives 6x = 30, so x = 5, and then 3y = 26 − 20 gives y = 2. So x − y = 3; the choice 7 is x + y.

Lines with different slopes cross exactly once, and only the last pair has different slopes (2 and −2). The first pair is parallel (no solution), and the other two pairs describe the same line twice (infinitely many solutions).

From s + l = 120 and 3s + 5l = 460, substituting s = 120 − l gives 2l = 100, so l = 50 and s = 70. The difference is 70 − 50 = 20; the choices 50 and 70 are the individual counts.

From the second equation, x = y − 1. Substituting gives 2y − 2 + 3y = 18, so y = 4 and x = 3, and a + b = 7.