Given the equation y=12x−32y=\frac12x-\frac32, which of the following ordered pairs is a solution?

( 4 , 1 2 ) \left(4,\frac12\right) ( 4 , 2 1 )

Given the equation y = 1 2 x − 3 2 y=\frac12x-\frac32 , which of the following ordered pairs is a solution?

Given the equation y equals one half x minus three halves, which of the following ordered pairs is the solution? So let's start by checking case a. So for case a, we have the x value of four and the y value of one half. Remember these ordered pairs are in the form x y. So let's go ahead and see if if this situation for point a, let's see if this is a solution to the equation. So that means taking my y's and replacing them with one half, then taking my x's and replacing them with four. So we'll get that y is one half. I'm just gonna replace that right there, and then we'll see that equals one half of x, which I can see is four, minus three halves. Now let's check this. We wanna see if these are equal. So we'll keep this one half on the outside here, and then one half of four is two. So we get two minus three halves. Now what I need to do is find some way to combine these into one fraction. Well, I can do that by getting like denominators. Now four is or two is the same thing as four over two, and then we'll have minus three over two. Now what I can do now that I have the same denominator is just combine the numerators four minus three. Four minus three is one. So we get that one half is equal to one half, and that is a true statement right about there. So we can say that case a is a solution to this problem. So this right here would be an ordered pair solution. Now rather than just stopping here, I'm gonna go ahead and check the other points as well, the other ordered pair. So I wanna see if there are any other cases where we have the solution. So let's now move to b. For b, we have two and one half, so let's check that up here. So take my y and replace it with one half because that's our y, and then we have that equals one half of two, replacing that with x, and then that's minus three halves. Now we keep the one half out front, and then we wanna see if this is equal to one half of two, which is one, minus three halves. Now one is the same thing as two over two, which will be minus three over two, and so I can just check the numerators giving us negative one, two minus three. So that one half is equal to negative one over two. And I can see that this is clearly not a true statement. One half does not equal negative one half. So therefore, b does not check out. Now we're going to check c. For c, we have the point six zero. So that means I'm going to take my y and replace it with zero, and we'll have that equals one half times an x value of six minus three halves. Let's see if these are equal. So we'll keep the zero on the left side, and one half of six is three, then we have three minus three halves. Now I can already tell these are not going to be equal since three and three halves are not the same, so this won't come out to zero. But let's go ahead and combine these anyways. So three is the same thing as six over two. So we'll write this as six over two because six divided by two is three minus three halves, and then subtracting the numerators, we end up getting just three halves. And clearly zero does not equal three halves, so point c does not check out. Now lastly, I'll move here to d for zero negative three. So that means taking my that means taking my y and replacing it with negative three, and then we'll set that equal to one half of zero because that's my x here, minus three halves. Let's go ahead and just work through this. So we have negative three upfront, one half of zero is zero, then we have minus three halves. We're checking if these are equal, and I can see that negative three does not equal negative three halves. So because of that, d does not check out either. Meaning a is the only solution to this problem. So this is how we can check to see if an ordered pair is a solution. As we can see, point a is the one that matches. Hope you found this video helpful, and let's move on.

Given the equation y = 1 2 x − 3 2 y=\frac12x-\frac32 , which of the following ordered pairs is a solution?

( 4 , 1 2 ) \left(4,\frac12\right) ( 4 , 2 1 )

( 2 , 1 2 ) \left(2,\frac12\right) ( 2 , 2 1 )

( 6 , 0 ) \left(6,0\right) ( 6 , 0 )

( 0 , − 3 ) \left(0,-3\right) ( 0 , − 3 )

4. Graphs and Functions

The Rectangular Coordinate System

Intermediate Algebra

4. Graphs and Functions

The Rectangular Coordinate System

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Multiple Choice

Given the equation $y = \frac{1}{2} x - \frac{3}{2}$ , which of the following ordered pairs is a solution?

$\left(4,\frac12\right)$

$\left(2,\frac12\right)$

$\left(6,0\right)$

$\left(0,-3\right)$

Verified step by step guidance

Start with the given equation: $y=\frac{1}{2}x - \frac{3}{2}$.

To check if an ordered pair $(x,y)$ is a solution, substitute the $x$-value from the pair into the equation to find the corresponding $y$-value.

Calculate the right side of the equation after substitution: $\frac{1}{2} \times x - \frac{3}{2}$.

Compare the calculated $y$-value with the $y$-value from the ordered pair. If they are equal, the pair is a solution to the equation.

Repeat this process for each given ordered pair to determine which one satisfies the equation.

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