Question

1. Solve by the linear combination method (with or without multiplication).

2x + 8y = –2
5x + 6y = 9 (Points : 1)
(–1, 3)
(3, –1)
(–2, 2)
(–2, 7)

Answer 1

ill help

Answer 2

thanks

Answer 3

Solve for e for the first equation.

e = -d + 2

Substitute that for the second equation and solve for d:

d(-d + 2) = 4
-d² + 2d = 4
-d² + 2d - 4 = 0

By quadratic formula, which is d = (-b² ± √(b² - 4ac))/2a:

a = -1
b = 2
c = -4

Substituting those values to the formula gives:

d = (-2 ± √(2² - 4(-1)(-4)))/2(-1)
d = (-2 ± √(4 - 16))/-2
d = (-2 ± √-12)/-2

Since the discriminant b² - 4ac gives nonreal solutions, there are no real solutions for the system of equations.

2. y = 3x - 7
6x - 2y = 12

Substitute y with 3x - 7 for the second equation:

6x - 2(3x - 7) = 12
6x - 6x + 14 = 12
0x + 14 = 12
0x = -2
x = undefined

Since 12 is not divisible by 0, x is not defined. Hence, there is also no solution for that system of equations.

3. 3x + 2y = 11
x - 2 = -4y

Solve for x for the second equation:

x = 2 - 4y

Substitute that for the first equation:

3(2 - 4y) + 2y = 11
6 - 12y + 2y = 11
6 - 10y = 11
-10y = 11 - 6
-10y = 5
y = -½

Substitute that for either equation:

x - 2 = -4(-½)
x - 2 = 2
x = 2 + 2
x = 4

Hence, the solution is (4,-½).

I hope this helps!

Answer 4

i haven't even reached quadratic equation yet

Answer 5

The linear combination method is basically the old elimination method

multiply each of the equations so that 1 term can be eliminated
then add the 2 equations.

looking at the problem I would multiply the 1st equation by 5 and the 2nd equation by -2

10x + 40y = -10
-10x -12y = -18
----------------- add the equations and x will be eliminated
you can do this bit

then solve for y.

when you get y substitute into either of the original equations to find x.

Answer 6

i was wondering of this correct?