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
$\frac{2x}{2} = \frac{-6y - 12}{2}$ < 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
$\frac{2y}{2} = \frac{-2}{2}$ (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
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I am not quite sure how to solve the second problem!

Answer 2

Answer:

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

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

Step-by-step 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 $\frac{a_1}{a_2},\frac{b_1}{b_2},\frac{c_1}{c_2}$

Thus, $\frac{a_1}{a_2}=\frac{2}{10}=\frac{1}{5}$

$\frac{b_1}{b_2}=\frac{6}{32}=\frac{3}{16}$

and $\frac{c_1}{c_2}=\frac{-12}{-62}=\frac{3}{8}$

hence, it is clear from above ratios, that $\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

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).