Math: e1248a5c

ID: e1248a5c

(SAT Suite Question Bank > Find Questions > Assessment: SAT + Test: Math + Domain: Algebra)

Comment: The difference between an infinite number of solutions and no solution is marginal, in terms of the cosmetic differences between systems of equations.

Method 1: An infinite number of solutions will have two identical equations in a system, so we just have to match. Because fractions are generally more challenging to work with than integers, why not multiply out the fractions? There is no need to put into slope-intercept form, y = mx + b.

(1/2)x + (1/3)y = 1/6

6[(1/2)x + (1/3)y = 1/6]

3x + 2y = 1

If the second equation, ax + y = c, matches the one above, then the two lines will be the same, but what makes this a Hard question is that the coefficient for y in the "alphabet soup" equation is 1, not 2. To make these two equations match, we could either manipulate the first equation above or the second. My preference is to leave the first one alone at this point, since we just worked it into having integer coefficients.

2(ax + y = c)

2ax + 2y = 2c

The question is asking for the value of a, so we can ignore the other parts of the equation. It must be true that

2ax = 3x

2a = 3

a = 3/2 or 1.5

The correct answer is (D).

On my scrap paper, I would probably write little more than 3x + 2y = 1 before selecting the answer. There is nothing wrong with writing out more steps, but as long as you keep the information organized, it is probably best to find the right combination of mental math and written numbers or equations that will allow you to work through the problem more efficiently.

Method 2: This question seems hand-selected for Desmos. Graph the first equation.

The second equation would have to match the first to yield an infinite number of solutions. However, if you want to see the solution for a, write a second equation, but leave the coefficient of y alone and set the equation equal to 1/6. Add a slider for a. This line will be parallel to the first (i.e. no solution), but that is only because we left y alone so that we could single out a.

There is no doubt that when a = 1.5, that is the only time these lines could become one.

Method 3: If you had a conceptual understanding of how two linear equations would yield an infinite number of solutions, but the second equation looked intimidating, you could plug in each answer choice and work through the algebra. Needless to say, this would not be the most efficient way to work through the problem, but it could save you in a pinch. Test (A), for instance.

(1/2)x + (1/3)y = 1/6

(-1/2)x + y = c

Adding the two equations would eliminate the x's. Remember, too, that that "ghost" coefficient in front of y in the second equation is 1, or, in terms of thirds, 3/3.

(1/3)y + (3/3)y = (1/6) + c

(4/3)y = (1/6) + c

y = (3/4)[(1/6) + c]

That would yield a single value for y, so (A) cannot be the correct answer. To skip ahead a bit, run through the same process with (D), 3/2.

(1/2)x + (1/3)y = 1/6

(3/2)x + y = c

Multiplying the top equation by 3 will make the coefficients for both x and y match in the two equations.

3[(1/2)x + (1/3)y = 1/6]

(3/2)x + (3/3)y = 3/6

(3/2)x + y = 1/2

If c = 1/2 in the equation a few lines above, then there would be an infinite number of ways x and y could combine to yield a sum of 1/2 (for instance, (0, 1/2) or (1/3, 0), to name just two). Again, I would not recommend solving the question this way, but it is functional.