MOST can’t solve this ALGEBRA Equation!
Вставка
- Опубліковано 6 вер 2024
- How to solve a non-linear system.
TabletClass Math Academy - TCMathAcademy....
Help with Middle and High School Math
Test Prep for High School Math, College Math, Teacher Certification Math and More!
Popular Math Courses:
Math Foundations
tabletclass-ac...
Math Skills Rebuilder Course:
tabletclass-ac...
Pre-Algebra
tabletclass-ac...
Algebra
tabletclass-ac...
Geometry
tabletclass-ac...
Algebra 2
tabletclass-ac...
Pre-Calculus
tabletclass-ac...
To start, there's 2 unknowns:
X and Y
And there's two equations:
X^2+Y^2 = 8 eq.1
X = (1/2)Y+1 eq.2
AND SINCE THERE'S ^2 power
means there's two solutions
to simplify:
2X = Y + 2 eq.2.1
Y = 2X - 2 eq.2.2
eq.2.2 into eq.1
X^2 +[2X-2]^2 = 8 eq.1.1
X^2 +4X^2 -8X +4 = 8
5X^2-8X - 4 = 0 eq1.2
quad equation:
[-b +/- sqrt(b^2-4ac)]÷2a
a = 5
b = -8
c = -4
[8 +/-sqrt(64+80)]÷10
[8+/-sqrt(144)]÷10
[8+/-(12)]÷10
20/10, -4/10
X.1 = 2
X.2 = -0.4
Y = 2X - 2 eq.2.2
Y.1 = 2(2)-2
= 2
Y.2 = 2(-0.4)-2
= -2.8
VERIFY:
X.1, Y.1 = (2,2)
X.2, Y.2 = (-0.4, -2.8)
c
X^2+Y^2 = 8 eq.1
2^2+2^2 = ? 8
4 + 4 =❤ 8✔️
(-0.4)^2 + (-2.8)^2 =? 8
0.16 + 7.84 =?8
8 =❤ 8✔️
FINAL ANSWER:
X.1, Y.1 = (2,2)
X.2, Y.2 = (-0.4, -2.8)
COMMENT:
X^2+Y^2 = 8 is a circle.
X = (1/2)Y + 1 is a straight line
the straight line intersects the circle in two places.
x= 2 and y = 2. I solved this in my head. Now, I'll watch to see if he has some other surprise for us. I didn't forsee 2 negative fractions being squared as another possible solution. Oh well, can't say that I even care about a second possible solution.
Looking at the first one I just thought that it would work if X and Y were both 2
2 squared + 2 squared = 4 + 4 = 8
and then, looking at the second one, I realised that that worked as well
Y = 1/2 X + 1 2 = 1/2(2) + 1 2 = 1 + 1
That's wut I did, took about 6 seconds.
Greetings. The answers are (2,2) and
(+,-14/5, -2/5). Based on the expressions given, using equation 2
gives X =2Y-2 We will now substitute this value of X in equation 1 to get
(2Y-2)^2 +Y^2=8 and
4Y^2-8Y+4+Y^2=8,
(4Y^2+Y^2)-8Y +4-8=0,
5Y^2-8Y-4=0, (5Y+2)(Y-2)=0.
Now, setting (5Y+2)=0, and (Y-2)=0,
we have Y=2, -2/5. We shall now substitute these values of Y in equation 2 and by doing so it is determined that X equals 2, -14/5. However, I have also determined that X equals 14/5 if -2/5 is substituted in equation 1. Therefore, when Y=-2/5, X=-14/5 or 14/5.
good one. very involved. thanks for the explanation.
y= -2/5, of course
Substitute y=1/2x +1 into y squared, solve for x, find x and substite back to find y!
Why when you factored this did you write the factors in different order? Isn’t it more logical to read it as (5x + 14) (x - 2)
This problem encompasses many algebraic equations to solve. And only got as far as (2,2). However John’s explanation made it look straightforward, as he usually does.
One aspect I didn’t understand however is the stand alone X in the calculation; x2 +1/4x2+x+1=8.
Otherwise enjoying John’s videos and comments in general.
Paul
I better double check when x= -14/5
I decided x=2 and y=2, but if the numbers were larger I wouldn't know how to go through all those steps.
Where’d the x come from in x^2+1/4x^2+x+1?
Bit late, but foil out (1/2X + 1) (1/2X + 1)
gives
1/4X^2 + 1/2X+ 1/2X +1
Which simplifies to
1/4X^2 + X +1
John cut out this part to the simplified answer for some brevity.
Thank you very much SIr
Thank you, after 60+ years I’m enjoying the review.
I struggled with this one
X=√5&8/9,Y=√3&8/9
X^2+y^2=8.....why did he decide that y =1/2x+1???
How is not x=2.5 y=1.5?
I also forgot x can=y
I forgot about the circle equation, been out of school since late70's
X=2,_2,8
Y=2,_0.4
I loved Maths at school but Id hated it if you'd been my teacher. Too many words and irrelavancies
You are not teach in details this way
I do not agree. This is a valid algebra problem By any means At least this time. I know this youtuber has a track record of ambiguity