solve the system using substitution: 3x^2 - 2y^2=1 4x - y = 3?

3x^2 - 2y^2=1

4x - y = 3

from second y=4x-3

substitute y in first equation

3x^2-2(4x-3)^2=1

3x^2-32x^2+48x-18-1=0

29x^2-48x+19=0

solve for x and result x1 and x2 then compute y1 and y2

System have 2 solutions because a straight line cut a hyperbola in 2 points in this case

B. If your system is

(is not clear if it's 14 0r 1 and 4 to second equations)

3x^2 - 2y^2=14

x - y = 3

then

3x^2-2(x-3)^2=14

x^2+12x-32=0

idem solve for x1 and x2 then y1,y2 .....

First, for between the equation resolve for x in terms of y or y in terms of x. right here i will resolve for y based on the 1st equation: y = 4x-3. Now plug that throughout to the 2d equation for y: 9x-2(4x-3)=10. Distribute: 9x-8x+6=10. Simplify: x=4. Now plug that into the two equation to resolve for y or to keep a while basically use the equation we derived for y interior the 1st step: y=4x-3=4(4)-3=sixteen-3=13. a answer might desire to fulfill the two equations. you could plug in x=4 and y=13 into the two equations to get actual statements, i.e. 3=3 and 10=10 after some simplifying. in spite of the shown fact that, utilising x=a million and y=a million satisfies the 1st equation yet no longer the 2d as a results of fact 7 would not equivalent 10. the respond is (4, 13). desire this helps!

3x^2 - 2y^2=1

4x - y = 3

Solve equation 1 for y.

-y=3-4x

y=-3+4x

Substitute -3+4x for y in equation 1.

3x^2 - 2y^2=1

3x^2 - 2(-3+4x)^2=1

Solve

3x^2+6-8x^2=1

9x+6-64x=1

-55x+ 6=1

-55x=-5

x=11

Substitute 11 for x in equation 1.

4x - y = 3

4(11) - y = 3

Solve

44-y=3

-y+44=3

-y=-41

y=41

The answer is (11, 41)