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For problems 1–4, solve the given system of equations using either substitution or elimination.

For Problems 14 Solve The Given System Of Equations Using Either Substitution Or Elimination class=

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Answer:

1. -3, -2

2. 2, -7

3. 1, 2.5

4. infinitely many solutions

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Answer:

1. (-3,-2)

2. (1, -7)

3. (1, 2.5)

4. No solution

Step-by-step explanation:

Number 1: Substitution

Plug y in to substitute and get

2x + (2x + 4) = -8

4x + 4 = -8

4x = -12

x = -3

Then plug into the y equation

y = 2(-3) + 4

y = -6 + 4

y = -2

(-3, -2)

Number 2: Elimination

Begin eliminating by crossing out same values

40x + 12y = -44

-2y = 14

Solve for y

-2y = 14

y = -7

Solve for x

40x + 12(-7) = -44

40x - 84 = -44

40x = 40

x = 1

(1, -7)

Number 3: Elimination

Add the two and get:

-3x + 2y = 2

6y = 15

Solve for y:

6y = 15

y = 5/2

Solve for x:

3x + 4(5/2) = 13

3x + 10 = 13

3x = 3

x = 1

(1, 2.5)

Number 4: Substitution

Solving for y to begin with:

4x - y = -3

-y = -3 - 4x

y = 3 + 4x

Then solve for y in the second system

-8x + 2y = 6

2y = 6 + 8x

y = 3 + 4x

Set equal to each other

3 + 4x = 4 + 4x

3 = 4

No solution