Question

1.)How many solutions are there to the following system of equations ?

2x+6y=-12
10x+32y=-62

0
2
1
infinitely many

2.) Solve the following system of equations by linear combination:
2d+e=8
d-e=4

The solution is (5,-2).
The solution is (4,0).
There is no solution.
There are an infinite number of solutions.

Answer 1

2x + 6y = − 12 ;10x + 32y = −62
Now make sure to solve!
2x + 6y = −12 for x > 2x + 6y + −6y = −12 + −6y (Just add -6y to both sides)
2x = −6y −12

`(2x)/(2) = (-6y - 12)/(2)` < Divide both sides by 2
x= − 3y − 6
Now we have to substitute > −3y − 6 for x in 10x + 32y = −62
10x + 32y = −62
10(−3y − 6) + 32y = −62
2y − 60 = −62 (We have to simplify both sides!)
2y − 60 + 60 = −62 + 60 (Now add 60 to both sides!)
2y = −2

`(2y)/(2) = (-2)/(2)` (Make sure to divide both sides by 2)
We get y = -1 here.
Substitute −1 for y in x = −3y − 6
−3y − 6x
(−3)(−1) − 6
x=−3 (Simplify both sides)
We have 2 solutions > x = −3 and y = −1
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am not quite sure how to solve the second problem!

Answer 2

1) Option (3) is correct. The system has one solution.

2) Option (2) is correct. The solution is (4,0)

Explanation:

1) Consider the given system of equation ,

2x +6y = -12 .....(1)

10x + 32y = -62 ....(2)

Here,
`a_1=2,b_1=6,c_1--12` and
`a_2=10,b_2=32,c_2=-62`

Find the ratio for
`(a_1)/(a_2),(b_1)/(b_2),(c_1)/(c_2)`

Thus,
`(a_1)/(a_2)=(2)/(10)=(1)/(5)`

`(b_1)/(b_2)=(6)/(32)=(3)/(16)`

and
`(c_1)/(c_2)=(-12)/(-62)=(3)/(8)`

hence, it is clear from above ratios, that
`(a_1)/(a_2)\\eq (b_1)/(b_2)\\eq(c_1)/(c_2)`

Thus, the system will have a unique solution.

We will solve the system ,

Multiply equation (1) by 5, we get,

10x + 30y = -60 ...(3)

Subtract equation (3) from (2),

10x + 32y -( 10x + 30y ) = -62-( -60 )

⇒ y = - 1

Put y = 1 in (1), we get,

2x +6(-1) = -12 ⇒ 2x= -12 +6 ⇒ x = -3.

Thus, (-3 , -1) is the solution

Hence the system has one solution.

2) Consider the given system of equation ,

2d + e = 8 ...........(1)

d - e = 4 ..........(2)

Add equation (1) and (2) , we get,

2d + e +(d -e) = 8+ 4

⇒ 2d +d = 12

⇒ 3d= 12

⇒ d = 4

Put d = 4 in (1) , we get

2d + e = 8 ⇒ 2(4)+e = 8 ⇒ e = 8 - 8 ⇒ e = 0.

Thus, The solution is (4,0).