My son needed to go to summer school, the teacher is an english teacher and doesn't know how to do the algerbra she is suposed to be teaching. So, he can't find any help with these last 2 problems and he needs them to finish the class. So if anyone can help it would be appreciated. Thank you!

A. Solve the quadratic-linear systems below algebraically

1. 3x - 1 + 2y = 0
3x^2-y^2+4 = 0


2. x + 2y = 4
y^2 - xy = 7

ok give me some time. I just did this stuff

here is #2
the Way I do it. teachers don't like it but it works... lol
You would have to do it this way because 7=prime number

LOL that is full marks in my school.

Out of curiosity what grade is your son in? I did this to my level Grade 12

Also I solved this as a polynomial (more than one variable) as a quadratic function.

ax^2+-bx+-c=0

Therfore hence ax^2+-bx+-c=0 and 2x^2-2x-3=0 it is solved

Thank you very much! I'm sure this going to help him. Its all greek to me LOL.

He's going into 12th grade in September.

He is the same age as me. then. are you sure you got the question right though? because 7 is a prime number. It would be almost impossible.

Jagr wrote:here is #2

Code: Select all

1. x+2y=4  Basic
2. x+y=2   Division
3. y=2-x    Move one of the variables :)

y^2-xy=7 Basic

4. y^2-xy=7                .You want to put the previous equation into the more complicated one.
5. (2-x)^2 -x(2-x)=7           .Now it looks like this! ;)
6. (2+x)(2-x)-2x+x^2=7       .Continue to solve  :roll: 
7. 4+2x-2x+x^2-2x+x^2=7    .Boring.... lol
8. x^2+x^2+2x-2x-2x+4-7=0       .Re-arrange.... again
9. 2x^2-2x-3=0                          .Where am I going again?.... Oh right. lol  That is Theoretically Correct. Your son should be able to solve it from there. ;)

The solution above is incorrect. step 2 is incorrect and so is step 7.
Jagr's method of solving the is problem is correct.

Here is the correct solution.

Given:
x + 2y = 4 -> x + 2y -4 = 0
y^2 - xy = 7 -> y^2 -xy - 7 = 0

Taking the 1st equation:
x = -2y + 4

By Substitution:
the 2nd equation becomes:
y^2 - (-2y + 4)y - 7 = 0

Multipiling:
y^2 - ( -2y^2 + 4y ) - 7 = 0

Remember the sign:
y^2 + 2y^2 - 4y - 7 = 0

Combine like terms:
3y^2 -4y - 7 = 0

Now the equation is in quadratic form:
3y^2 -4y - 7 = 0

There is now a couple of ways you can get to the roots.

The brainiac :
Mutiply the equation by 1/3:
1/3 ( 3y^2 - 4y - 7 = 0 )

Giving:
y^2 - 4/3y - 7/3 = 0

Now your roots will be in the form:
( y +/- ? )(y +/- ?? )

WHERE
? * ?? is the negative number -7/3 so either ? or ?? must be negative
so ( y + ? )(y - ??)
AND
? + ?? is the negative number -4/3

The combination of ? and ?? is:
( y + 1 )( y - 7/3 )

Check by expansion:
y*y - 7/3y + y -7/3

note: y = 3/3y

Combine like terms:
y^2 -4/3y -7/3 = 0

Therefore:
y1 = -1 , y2 = 7/3



Alternate Way:
The Quadratic formula:
y1 = 1/2a ( -b + SquareRoot( b^2 - 4ac )
y2 = 1/2a ( -b - SquareRoot( b^2 - 4ac )

The equation in quadratic form:
ay^2 + by + c = 0
3y^2 - 4y - 7 = 0

y1 = 1/2*3 ( -( -4 ) + SquareRoot( (-4)^2 - (4 * 3 * -7) )
y1 = 1/6 ( 4 + SquareRoot ( 16 - ( -84 ) )
y1 = 1/6 ( 4 + SquareRoot( 16 + 84 ) )
y1 = 1/6 ( 4 + SquareRoot( 100 ) )
y1 = 1/6 ( 4 + 10 )
y1= 7/3

y2 = 1/6 ( -( -4 ) - SquareRoot( 100 ) )
y2 = 1/6 ( 4 -10 )
y2 = 1/6 (-6 )
y2 = -1


Now back to the Original Equations:
x + 2y -4 = 0
y^2 -xy - 7 = 0

Taking the 1st equation:
When y = y1
then:
x1 + 2( 7/3 ) - 4 = 0
x1 + 14/3 -4 = 0
x1 = -14/3 + 4
x1 = -14/3 + 12/3
x1 = -2/3

When y = y2
then:
x2 + 2( -1 ) -4 = 0
x2 -2 -4 = 0
x2 = 2 + 4
x2 = 6

Therefore the solution set of the quadratic system:
x + 2y -4 = 0
y^2 -xy - 7 = 0

is:
(-2/3, 7/3), (6, -1)

Problem 1

Given:
3x - 1 + 2y = 0
3x^2 -y^2 + 4 = 0

Taking Eq.1
2y = -3x + 1
y = -3/2x + 1/2

Substituting in Eq.2
3x^2 - (-3/2x + 1/2)^2 + 4 = 0

Expanding:
3x^2 - [(-3/2x + 1/2)(-3/2x + 1/2)] + 4 = 0

Multipling:
3x^2 - [ 9/4x^2 -3/4x -3/4x +1/4 ] + 4 = 0

Remember the sign:
3x^2 -9/4x^2 +3/4x +3/4x -1/4 + 4 = 0

Combining like terms:
12/4x^2 -9/4x^2 +3/4x +3/4x -1/4 +16/4 = 0
3/4x2 + 6/4x + 15/4 = 0

Multipling to simplify:
4/3 ( 3/4x^2 + 6/4x + 15/4 = 0 )

The quadratic form:
ax^2 + bx + c = 0
x^2 + 2x + 5 = 0

The roots:
( x +/- ? )(x +/- ??)

Use the quadratic formula:
1/2a( -b +/- SquareRoot( b^2 - 4ac ))


1/2 ( -2 +/- SquareRoot( 2^2 - ( 4 * 1 * 5 ) )
1/2 ( -2 +/- SquareRoot( 4 - 20 )
1/2 ( -2 +/- SquareRoot( -16 )

since no real numbers may satisfy the SquareRoot, the roots are imaginary numbers,
where i^2 = -1 and SquareRoot( -1 ) = i
so:
1/2 ( -2 +/- 4i ) = -1 +/- 2i

x1 = -1 + 2i
x2 = -1 - 2i

Returning to the original Equations:
3x + 2y - 1 = 0
3x^2 - y^2 + 4 = 0

Substituting in Eq.1
where x1 = x
3( -1 + 2i) + 2y - 1 = 0
-3 + 6i + 2y - 1 = 0
2y = 3 + 1 - 6i
y = 2 - 3i

where x2 = x
3( -1 - 2i) + 2y - 1 = 0
-3 -6i + 2y - 1 = 0
2y = 3 + 1 + 6i
y = 2 + 3i

Therefore the solution set for the quadratic system:
3x + 2y - 1 = 0
3x^2 - y^2 + 4 = 0
is:
( -1 +2i, 2 - 3i ), ( -1 - 2i, 2 + 3i )

2+2=4

iP0kE do good?

Oh my god, WOW! thats amazing stuff, I don't recall doing anything like that back in my days of schooling. I liked when my kids where in grade school much better and I could actually help them with their homework. I can't thank you enough sir!

My name is Derring , and I'm not smarter than a 5th grader.