find the coordinates of the points at which the line with equation y= 3x - 1 intersects curve with equation y^2 - xy = 15

Question

simultaneous sum

Answer 1

Hi Rodela,

y=3x-1------Eq-1

y^2-xy=15-------Eq2

Substitute the value of y from eq 1 to Eq 2,

we get,

(3x-1)^2-x(3x-1)=15

9x^2-6x+1-3x^2+x=15

6x^2 -5x+1-15=0

6x^2-5x-14=0

6x^2-12x+7x-14=0

6x(x-2)+7(x-2)=0

(x-2)(6x+7)=0

x=2 or x= -7/6

now if x=2, y= 3(2)-1=5

  if x=-7/6, y=3(-7/6)-1=-7/2-1=-9/2.

Ans:  (2,5) & (-7/6,-9/2)

 

Meena from strongsville,OH

 

Answer 2

The equation y^2 -x y = 15 is a hyperbola tilted with respect to the coordinate axes.  The other equation is a straight line.   There will be two points of intersection.    The easiest analytic approach is to solve the straight line equation for x in terms of y:   x =  y/3 + 1/3.    This expression can then be substituted into the hyperbola equation to get an equation for y by itself.   The resulting equation is parabolic.   It can be rearranged to the standard form:

 

2 y² - y -15 = 0

 

The left hand side factors   (2 y +9 ) (y-5) = 0   so the solution values of y are  -9/2  and 5.

 

The corresponding x values can be obtained by substitution into the straight line equation.  This gives us the coordinates of the two points of intersection

 

(- 7/6, -9/2)    and   (2,5)

 

Answer 3

Hi Rodela;

y= 3x - 1

y^2 - xy = 15

Let's take the first equation and replace y of the second equation with (3x-1)

[(3x-1)²]-[x(3x-1)]=15

Let's FOIL the first bracketed equation...

FIRST...(3x)(3x)=9x²

OUTER...(3x)(-1)=-3x

INNER...(-1)(3x)=-3x

LAST...(-1)(-1)=1

9x²-3x-3x+1

9x²-6x+1

 

Let's distribute the - and the x in the second bracketed equation...

-3x²+x

 

9x²-6x+1-3x²+x=15

Combine like terms...

6x²-5x+1=15

Subtract 15 from both sides...

6x²-5x+1-15=15-15

6x²-5x-14=0

For the FOIL...

FIRST must be (6x)(x) or (3x)(2x)=6x²

OUTER and INNER must add-up to -5x.

LAST must be (14)(1) or (1)(14) or (2)(7) or (7)(2) and one number must be negative while the other positive to render the sum of -5 and product of -14.

(6x+7)(x-2)=0

Let's FOIL...

FIRST...(6x)(x)=6x²

OUTER...(6x)(-2)=-12x

INNER...(7)(x)=7x

LAST..(7)(-2)=-14

6x²-12x+7x-14=0

6x²-5x-14=0

(6x+7)(x-2)=0

Either or both parenthetical equation(s) must equal zero...

6x+7=0, 6x=-7, x=-7/6

x-2=0, x=2

 

Let's return to the original parabolic equation...

y^2 - xy = 15

x=-7/6

y²-[(-7/6)y]=15

y²+7/6y=15

Let's eliminate the annoying fraction by multiplying both sides by 6...

6(y²+7/6y)=(15)(6)

6y²+7y=90

6y²+7y=90

6y²+7y-90=0

(3y-10)(2y+9)=0

Either or both parenthetical equation(s) must equal zero...

3y-10=0, 3y=10, y=10/3

2y+9=0, 2y=-9, y=-9/2

 

Original linear equation...

y=3x-1

y=[(3)(-7/6)]-1

y=(-21/6)-1

y=(-7/2)-1

y=(-7/2)-(2/2)

y=-9/2

 

(-7/6,-9/2)

 

Original parabolic equation...

y^2 - xy = 15

x=2

y²-2y=15

y²-2y-15=0

(y-5)(y-3)=0

y=5 or 3

Original linear equation...

y=3x-1

y=[(3)(2)]-1

y=6-1

y=5

 

(2,5)