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## SAT

### Course: SAT > Unit 10

Lesson 2: Passport to advanced mathematics- Solving quadratic equations — Basic example
- Solving quadratic equations — Harder example
- Interpreting nonlinear expressions — Basic example
- Interpreting nonlinear expressions — Harder example
- Quadratic and exponential word problems — Basic example
- Quadratic and exponential word problems — Harder example
- Manipulating quadratic and exponential expressions — Basic example
- Manipulating quadratic and exponential expressions — Harder example
- Radicals and rational exponents — Basic example
- Radicals and rational exponents — Harder example
- Radical and rational equations — Basic example
- Radical and rational equations — Harder example
- Operations with rational expressions — Basic example
- Operations with rational expressions — Harder example
- Operations with polynomials — Basic example
- Operations with polynomials — Harder example
- Polynomial factors and graphs — Basic example
- Polynomial factors and graphs — Harder example
- Nonlinear equation graphs — Basic example
- Nonlinear equation graphs — Harder example
- Linear and quadratic systems — Basic example
- Linear and quadratic systems — Harder example
- Structure in expressions — Basic example
- Structure in expressions — Harder example
- Isolating quantities — Basic example
- Isolating quantities — Harder example
- Function notation — Basic example
- Function notation — Harder example

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# Linear and quadratic systems — Basic example

Watch Sal work through a basic Linear and quadratic systems problem.

## Want to join the conversation?

- my ti84 only shows (16,5) co-ordinate :/(0 votes)
- Yeah not every problem will allow a calculator. My reliance on a calculator came back to bite me, too.(16 votes)

- plugging in the answers is probably the best and fastest method to do this math(7 votes)
- oh this question is SNEAKY! trying to trick me and preventing me from getting more POINTS!(5 votes)
- I think substitution works better with this problem + it is easier!(3 votes)
- i can't understand this method

is there any else(3 votes) - Or simple, use the elimination method(2 votes)
- Substitution is more efficient in this case as x is already isolated.(2 votes)

- can anyone who explain me why is nt A?(2 votes)
- Let's plug in the numbers and see if it works:

2y + 6 = x

y^2 - 9 = x

I'll substitute x = -5 and y = 3 in the system of equations above.

2 * 3 + 6 = -5 --> 12 =/= -5

9 - 9 = -5 --> 0 =/= -5

Neither equations work with the (-5, 3) pair, so A is definitely not the answer.(1 vote)

- I understood the ending clearly and easily but the starting is too complicated(2 votes)
- wouldn't it be y=-5 and y=3(1 vote)
- If you have y-5=0, and you plug in -5, it'll be -10, not 0. You add 5 to both sides, which means y=5

y+3=0 is the same thing. Subtract each side by 3 and you'll get y=-3.(2 votes)

- As another example, could anybody show me how the nonlinear expression

x-7y=25

x^2+y^2=25 could be solved using the substitution method?(1 vote)

## Video transcript

- [Teacher] Which of
the following represents all solutions X comma Y to the system of equations shown below? This is an interesting system of equations because this is a linear
equation, this first one, but the second one is nonlinear. You have a Y squared right over here. Well this one actually can
be solved with substitution because 2y plus six needs to be equal to X but then we also that X is
equal to Y squared minus nine. So we can take Y squared minus
nine and substitute it for X. So if you do that, you
then can solve for Y, the Y of a solution to this system. So if 2y plus six is equal to Y squared minus nine. And so let's see, let's just get zeroes
on the left hand side. So we could subtract 2y from both sides. I'll write that subtract 2y there. And we could subtract six from both sides. And then on the left hand side we're gonna be left with zero. And on the right hand
side you're gonna have Y squared minus 15 minus, actually let me
write it the other way, Y squared, let me write the minus 2y minus 15. And now to solve for the Ys that would satisfy this
quadratic right over here, we could factor this quadratic, we could factor this quadratic expression. And let's see, the way I
would do that is first of all kind of think of two numbers
whose product is negative 15. And immediately the
numbers three and five, either negative three, positive five or positive three, negative five jump out; and whose sum is negative two. Well if the sum is negative that means that the negative number has
to have a larger magnitude. So it's gonna be negative five and three. Negative five and three's
product is negative 15; their sum is negative two. And if what I'm doing right now looks a little bit like voodoo to you, I encourage you to review
factoring quadratic expressions on Kahn Academy. Do a search on Kahn Academy for factoring quadratic expressions. But we now know how to factor, this is gonna be zero is
equal to, we can write this as Y minus five times Y plus three. Well, if you have the
product of two things that are equal to zero that means that you could get this to equal zero by making one or both
of these equal to zero. So the solutions are gonna be how do you make this one equal zero? Well Y would be equal to five. And how do you make
this one equal to zero? Well Y would be equal to negative three. Either one of these
would make one of these, Y equals five would make this zero, which would make the
entire expression zero, so this is a solution. And Y equals negative
three would make this zero, which would make this
entire expression zero, so this is also a solution. Now we could go back and
try to solve for the Xs or we can immediately look at our choices and say well are either of these choices have Y equals five and
Y equals negative three as at least the
Y-coordinates of solutions. And these first two don't
even have two choices and but this one right over here, this one is actually trying
to trick us a little bit because you know we have this number five and negative three, it's like hey, maybe this one
but this isn't Y equals five, this is X equals five. And when X equals five, Y
is equal to negative three. That's X and that's Y. The first coordinate is X,
the second coordinate is Y. So Y equals five and Y
equals negative three. This one is Y is equal to five and Y is equal to negative three. So you should feel
immediately pretty good about this choice right over here. This one puts the five
and the negative three in for the X-coordinate. That says X is equal to five and X is equal to negative three. So if I was under a time pressure, I would just stop there and I would move on to the next question 'cause I feel good about this choice. But if you want to go back and feel good that when Y equals five, X equals 16 and Y equals negative
three, X is equal to zero, you can go back and substitute
into this first equation. Because we have two times five plus six is equal to X. So when Y is equal to five, you have 10 plus six. You have X is equal to 16,
when Y is equal to five. And that's what we saw right over here with when Y is five, X is 16. And when Y is negative three, two times negative three
plus six is equal to X. This on the left hand side is gonna be negative six plus six. X is equal to zero. So when Y is negative
three, X is equal to zero. So we took a little extra time to make sure that we feel good about this. But once again, under time pressure I wouldn't have necessarily gone through to this extra length.