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1. [tex]\( 2x + y = 3 \)[/tex]
2. [tex]\( x - 2y = -1 \)[/tex]

If equation 1 is multiplied by 2 and then the equations are added, the result is:

A. [tex]\( 3x = 5 \)[/tex]

B. [tex]\( 5x = 5 \)[/tex]

C. [tex]\( 3x = 2 \)[/tex]


Sagot :

Let's solve the given system of linear equations step-by-step in order to find the solution:

1. The system of equations is:

[tex]\[ \begin{cases} 2x + y = 3 & \text{(Equation 1)} \\ x - 2y = -1 & \text{(Equation 2)} \end{cases} \][/tex]

2. First, we will multiply Equation 1 by 2 to facilitate the elimination method. By multiplying Equation 1 by 2, we get:

[tex]\[ 2 \cdot (2x + y) = 2 \cdot 3 \][/tex]

This simplifies to:

[tex]\[ 4x + 2y = 6 \quad \text{(Equation 3)} \][/tex]

3. Now, we will add Equation 3 to Equation 2:

[tex]\[ (4x + 2y) + (x - 2y) = 6 + (-1) \][/tex]

4. Combining like terms on the left-hand side, we get:

[tex]\[ 4x + x + 2y - 2y = 6 - 1 \][/tex]

This further simplifies to:

[tex]\[ 5x = 5 \][/tex]

5. Solving for [tex]\(x\)[/tex], we divide both sides by 5:

[tex]\[ x = 1 \][/tex]

6. Now that we have [tex]\(x = 1\)[/tex], we substitute this value back into Equation 1 to find [tex]\(y\)[/tex]:

[tex]\[ 2(1) + y = 3 \][/tex]

This simplifies to:

[tex]\[ 2 + y = 3 \][/tex]

Solving for [tex]\(y\)[/tex], we subtract 2 from both sides:

[tex]\[ y = 1 \][/tex]

7. Therefore, the solution to the system of equations is:

[tex]\[ x = 1, \quad y = 1 \][/tex]
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