1. Solve the system by elimination.

x + 5y - 4z = -10
2x - y + 5z = -9
2x - 10y - 5z = 0
A. (5, –1, 0) B. (–5, 1, 0) C. (–5, –1, 0)** D. (–5, –1, –2)

2. Solve the system by substitution.
2x – y + z = -4
z = 5
-2x + 3y - z = -10
A. (-8, 7, 5)** B. (-8, -7, 5) C. (8, –7, 5) D. (-8, -7, -5)

3. A food store makes a 11-lb mixture of peanuts, almonds, and raisins. The cost of peanuts is $1.50 per pound, almonds cost $3.00 per pound, and raisins cost $1.50 per pound. The mixture calls for twice as many peanuts as almonds. The total cost of the mixture is $21.00. How much of each ingredient did the store use?
A) 3 lbs. peanuts, 6 lbs. almonds, 2 lbs. raisins B) 8 lbs. peanuts, 1 lb. almonds, 2 lbs. raisins C) 6 lbs. peanuts, 3 lbs. almonds, 2 lbs. raisins** D) 8 lbs. peanuts, 2 lbs. almonds, 1 lb. raisins

Someone pls check my answers ASAP. I'm very lost! Thanks :)

4 answers

It's

C
B
C
100%

1 is wrong, 2&3 remain correct

1. To solve the system of equations by elimination, the goal is to eliminate one variable at a time. Let's start by eliminating the "x" variable.

Looking at the given equations:
x + 5y - 4z = -10 ---(1)
2x - y + 5z = -9 ---(2)
2x - 10y - 5z = 0 ---(3)

We can see that equation (3) is already multiplied by 2, so it will be convenient to start by eliminating the "x" variable using equations (2) and (3).

Multiply equation (2) by 2:
2(2x - y + 5z) = -9
4x - 2y + 10z = -18 ---(4)

Now, subtract equation (4) from equation (3) to eliminate "x":
(2x - 10y - 5z) - (4x - 2y + 10z) = 0 - (-18)
2x - 10y - 5z - 4x + 2y - 10z = 18
-2x - 8y - 15z = 18 ---(5)

Next, to eliminate the "x" variable, subtract equation (5) from equation (1):
(x + 5y - 4z) - (-2x - 8y - 15z) = -10 - 18
x + 5y - 4z + 2x + 8y + 15z = -28
3x + 13y + 11z = -28 ---(6)

We now have two equations left:
3x + 13y + 11z = -28 ---(6)
-2x - 8y - 15z = 18 ---(5)

To proceed with elimination, let's eliminate the "x" variable again. Multiply equation (6) by 2:
2(3x + 13y + 11z) = -28
6x + 26y + 22z = -56 ---(7)

Now, subtract equation (7) from equation (5) to eliminate "x":
(-2x - 8y - 15z) - (6x + 26y + 22z) = 18 - (-56)
-2x - 8y - 15z - 6x - 26y - 22z = 18 + 56
-8x - 34y - 37z = 74 ---(8)

We have obtained a new equation:
-8x - 34y - 37z = 74 ---(8)

Finally, let's solve the system of equations (6) and (8) using either substitution or elimination method.

Multiply equation (8) by 3:
3(-8x - 34y - 37z) = 3(74)
-24x - 102y - 111z = 222 ---(9)

Now, add equation (6) and equation (9) to eliminate "x":
(3x + 13y + 11z) + (-24x - 102y - 111z) = -28 + 222
3x + 13y + 11z - 24x - 102y - 111z = 194
-21x - 89y - 100z = 194 ---(10)

We now have two new equations:
-21x - 89y - 100z = 194 ---(10)
-8x - 34y - 37z = 74 ---(8)

Let's solve this system of equations to find the values of x, y, and z.
-21x - 89y - 100z = 194 ---(10)
-8x - 34y - 37z = 74 ---(8)

You can use a system of equations solver, such as matrix methods or substitution, to find the values of x, y, and z.