System of Equations

Question

kx - 2y = 3

7x + 4y = 4

In the system of equations above, k is a constant and x and y are variables. For what value of k will the system of equations have no solution?

A) -21/4

B) -7/2

C) 7/2

D 21/4

Answer 1

Note that speed is important on the SAT. You need to give correct answers, and give them as fast as humanly possible.

When a system of linear equations has no solution, this means that two lines are parallel. Therefore, they have the same slope but a different y-intercept.

The linear equations are given to you in the standard form (also called general form). I wouldn't convert them to the slope-intercept form, but work with them as they are. Here is the standard form:

Ax+By=C

In this form, the slope is:

m=-A/B

The y-intercept is: C/B

(I would commit this theory to memory.)

kx - 2y = 3

m₁ = -k/-2 = k/2

7x + 4y = 4

m₂ = -7/4

m₁ = m₂

k/2 = -7/4

k = -7/2

So the answer is B.

Further considerations:

1. I wouldn't bother checking that the y-intercepts are unequal. I've never seen a SAT question that specified that there was no solution to a system, but then when you check there is actually in infinity of solutions. (Which would be the only other option here, given that the slopes are equal.)
2. On an actual test, I'd like to minimize the use of pen and paper. Because I know the theory, at most, I'd personally write: -k/-2 = k/2=-7/4 and perform the last step mentally.

Answer 2

You’re given this system of equations:

1. kx - 2y = 3
2. 7x + 4y = 4

The question asks: For what value of k will this system have no solution?

Step 1: Understand what "no solution" means.

When a system of equations has no solution, it means the lines are parallel. That means they have the same slope but different y-intercepts. So, to find the value of k that makes this happen, we need the slopes of both lines to be the same.

Step 2: Solve both equations for y to find their slopes.

Start with equation 1:

kx - 2y = 3

Subtract kx from both sides:

-2y = -kx + 3

Now divide by -2:

y = (k/2)x - 3/2

So the slope of the first line is (k/2)

Now look at equation 2:

7x + 4y = 4

Subtract 7x from both sides:

4y = -7x + 4

Now divide by 4:

y = (-7/4)x + 1

So the slope of the second line is -7/4

Step 3: Set the slopes equal to find the value of k.

To make the lines parallel, the slopes must be equal. So:

k/2 = -7/4

Now solve for k. Multiply both sides by 2:

k = -7/2

Final Answer: B) -7/2

Tip: When you're asked for when a system has "no solution," think about parallel lines. Set the slopes equal, solve for the variable, and make sure the lines have different y-intercepts. That's how you know they'll never intersect!

Let me know if you want help with more SAT system of equations problems — I'm happy to explain more!

Answer 3

My approach was somewhat different.

We are interested in a value for k such that the system has no solution.

We can simply start solving the system as we would normally solve for 2 variables - eliminate one variable.

In this case, eliminate y, by multiplying the first equation by 2:

2kx - 4y = 6,

and adding the two equations.

This leaves us with:

(2k + 7) x = 10

The RHS doesn't matter, because this equation is not solvable when 2k + 7 = 0, which gives us k = -7/2.

Thus, answer B is correct.

Answer 4

The system of equations will have no solutions when the two lines are parallel. The slopes will be the same but the y -intercepts will be different. In other words, the coefficients of both x variables will be the same value.

To find the value of k rewrite the linear equations in slope y - intercept form.

So kx -2y = 3

-kx -kx

-2y = - kx+3

-2 -2

y= kx - 3

2 2

And 7x + 4y = 4

-7x -7x

4y = -7x + 4

4. 4

y = -7 + 1

4

So setting the slopes equal to each other, we get:

K = -7

2 4

Cross multiply: 4k = -14

Divide both sides by 4: k = -14

4

k = -7

2

So the correct answer is (B).

Answer 5

This is a system of two linear equations. That system will not have a solution when the graphs of those equations (lines) are parallel. That happens when the slopes are equal and their y-intercepts are different. Note that if the slopes are the same and the y-intercepts are equal the equations are identical, so the system would have infinitely many solutions.

Transform each equation into slope-intercept form.

kx - 2y = 3

kx - 3 = 2y

y = (k/2)x - 3/2

The slope of this line is represented by k/2 and its y-intercept is at (0, -3/2).

7x + 4y = 4

4y = -7x + 4

y = (-7/4)x + 1

The slope of this line is -7/4 and its y-intercept is at (0,1).

The slopes will be equal (and lines parallel) when:

-7/4 = k/2

4k = -14

k = -14/4 = -7/2

When k = -7/2 the slope of the 1st line will be (1/2)(-7/2) = -7/4, which means the two lines are parallel with different y-intercepts, so the system has no solution.

Visit the following graph, use the slider on k to see different forms for the 1st equation. Then click the set link to make the lines parallel.

desmos.com/calculator/krtlk50qld

Answer 6

A system has no solutions when the ratios of the coefficients of x and y are equal, but the constant term ratio is different.

a₁x + b₁y = c₁

a₂x + b₂y = c₂

So, if (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂), then there is no solution.

In this example, we want to start by creating the ratio. -2/4 = -1/2. Then, we set (k/7) = -1/2 and arrive at k = -7/2. The ratios of the coefficients of x and y are now equal.

Now we check if the constant term ratio is different.

(c₁/c₂) = 3/4 ≠ -1/2.

Because the constant term ratio is different, it means that the lines are parallel but are not the same. Therefore, there is no solution.

The correct answer is indeed B) -7/2.