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Solve the simultaneous equations:

[tex]\[
\begin{array}{l}
2x + 3y = 13 \\
4x - y = -2
\end{array}
\][/tex]


Sagot :

To solve the system of simultaneous equations:

[tex]\[ \begin{cases} 2x + 3y = 13 \\ 4x - y = -2 \end{cases} \][/tex]

we can follow these steps:

### Step 1: Solve one of the equations for one variable

Let's solve the second equation for \(y\):

[tex]\[ 4x - y = -2 \][/tex]

Isolating \(y\):

[tex]\[ y = 4x + 2 \][/tex]

### Step 2: Substitute the expression for \(y\) into the first equation

Take the expression for \(y\) from the second equation and substitute it into the first equation:

[tex]\[ 2x + 3(4x + 2) = 13 \][/tex]

### Step 3: Simplify the substituted equation and solve for \(x\)

Distribute and simplify:

[tex]\[ 2x + 12x + 6 = 13 \][/tex]

Combine like terms:

[tex]\[ 14x + 6 = 13 \][/tex]

Subtract 6 from both sides:

[tex]\[ 14x = 7 \][/tex]

Divide both sides by 14:

[tex]\[ x = \frac{7}{14} = \frac{1}{2} \][/tex]

### Step 4: Substitute the value of \(x\) back into the expression for \(y\)

Now substitute \(x = \frac{1}{2}\) back into the expression for \(y\):

[tex]\[ y = 4 \left( \frac{1}{2} \right) + 2 \][/tex]

Multiply:

[tex]\[ y = 2 + 2 = 4 \][/tex]

### Step 5: Verify the solution

It's always a good idea to verify the solution by plugging \(x = \frac{1}{2}\) and \(y = 4\) back into the original equations to ensure they hold true:

For the first equation:

[tex]\[ 2 \left( \frac{1}{2} \right) + 3(4) = 1 + 12 = 13 \][/tex]

For the second equation:

[tex]\[ 4 \left( \frac{1}{2} \right) - 4 = 2 - 4 = -2 \][/tex]

Both equations are satisfied.

Therefore, the solution to the system of equations is:

[tex]\[ x = \frac{1}{2}, \quad y = 4 \][/tex]