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A. Solve the following system of equations using the substitution method and the elimination method.

1. [tex]\[ x + 2y + 1 = 0 \][/tex]
[tex]\[ 2x - 3y - 12 = 0 \][/tex]

2. [tex]\[ 2x - y - 3 = 0 \][/tex]
[tex]\[ 4x + y - 3 = 0 \][/tex]


Sagot :

Sure, let's solve these systems of equations step by step using both the substitution method and the elimination method.

### System 1: [tex]\(x + 2y + 1 = 0\)[/tex] and [tex]\(2x - 3y - 12 = 0\)[/tex]

#### Substitution Method:

1. Solve the first equation for [tex]\(x\)[/tex]:
[tex]\[ x + 2y + 1 = 0 \implies x = -2y - 1 \][/tex]

2. Substitute [tex]\(x = -2y - 1\)[/tex] into the second equation:
[tex]\[ 2(-2y - 1) - 3y - 12 = 0 \][/tex]

3. Simplify and solve for [tex]\(y\)[/tex]:
[tex]\[ -4y - 2 - 3y - 12 = 0 \implies -7y - 14 = 0 \implies y = -2 \][/tex]

4. Substitute [tex]\(y = -2\)[/tex] back into [tex]\(x = -2y - 1\)[/tex]:
[tex]\[ x = -2(-2) - 1 = 4 - 1 = 3 \][/tex]

So, the solution to the first system is:
[tex]\[ (x, y) = (3, -2) \][/tex]

### System 2: [tex]\(2x - y - 3 = 0\)[/tex] and [tex]\(4x + y - 3 = 0\)[/tex]

#### Elimination Method:

1. Add the two equations together to eliminate [tex]\(y\)[/tex]:
[tex]\[ (2x - y - 3) + (4x + y - 3) = 0 \implies 6x - 6 = 0 \implies 6x = 6 \implies x = 1 \][/tex]

2. Substitute [tex]\(x = 1\)[/tex] into the first equation to solve for [tex]\(y\)[/tex]:
[tex]\[ 2(1) - y - 3 = 0 \implies 2 - y - 3 = 0 \implies -y - 1 = 0 \implies y = -1 \][/tex]

So, the solution to the second system is:
[tex]\[ (x, y) = (1, -1) \][/tex]

### Summary of Solutions:
- First system: [tex]\((x, y) = (3, -2)\)[/tex]
- Second system: [tex]\((x, y) = (1, -1)\)[/tex]

These are the solutions for the given systems of equations using the substitution and elimination methods.