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Jordan is solving this system of equations: y = 2x2 + 3 and y – x = 6. Which statements are true about Jordan’s system? Check all that apply. The quadratic equation is in standard form. Using substitution, the system of equations can be rewritten as 2x2 – x – 3 = 0. There are two real number solutions. There are no real number solutions. A solution of the system of equations is (–1, 1.5).

2 Answers

2 votes

Answer:

its ABC

Step-by-step explanation:

User Eric Kim
by
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3 votes

Three Answers: choice A, choice B, choice C.

Basically the first three items are true, with the rest false.

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Step-by-step explanation:

Let's go through each answer choice one by one to see if it is true or false.

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A. The quadratic equation is in standard form.

This is true. Standard form for quadratics is y = ax^2+bx+c. In this case, a = 2, b = 0, c = 3. It might help to write the equation as y = 2x^2+0x+3.

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B. Using substitution, the system of equations can be rewritten as 2x2 – x – 3 = 0.

This is true. Start with y-x = 6. Replace y with 2x^2+3 and we get

2x^2+3-x = 6

From here we subtract 6 from both sides

2x^2+3-x-6 = 6-6

2x^2-x-3 = 0

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C. There are two real number solutions.

This is true. When you graph the two equations (I recommend GeoGebra but Desmos is another good tool), you'll see they intersect at two different locations. Each location is a solution in the form (x,y).

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D. There are no real number solutions.

This is false. It contradicts choice C.

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E. A solution of the system of equations is (–1, 1.5).

This is false. The two solutions are (-1, 5) and (1.5, 7.5) which is where the two graphs intersect.

Another way to check is to plug x = -1 into either equation. You'll find the result to both is y = 5 instead of y = 1.5

User Machiel
by
4.9k points
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