How many solutions does this system have?

[tex]\[
\left\{
\begin{array}{r}
2x + 8y = 16 \\
-3x + 6y = 30
\end{array}
\right.
\][/tex]



Answer :

To determine the number of solutions for the given system of linear equations, we will analyze the equations step by step.

The system of equations is:
[tex]\[ \begin{cases} 2x + 8y = 16 \quad \text{(1)}\\ -3x + 6y = 30 \quad \text{(2)} \end{cases} \][/tex]

First, let’s simplify each equation, if possible.

Equation (1):

[tex]\[ 2x + 8y = 16 \][/tex]

Divide every term by 2 to simplify:

[tex]\[ x + 4y = 8 \quad \text{(3)} \][/tex]

Equation (2):

[tex]\[ -3x + 6y = 30 \][/tex]

Divide every term by 3 to simplify:

[tex]\[ -x + 2y = 10 \quad \text{(4)} \][/tex]

Now let’s observe the simplified equations (3) and (4):

[tex]\[ \begin{cases} x + 4y = 8 \quad \text{(3)}\\ -x + 2y = 10 \quad \text{(4)} \end{cases} \][/tex]

To solve for \(x\) and \(y\), we can use the method of elimination or substitution. Here, we'll use elimination to eliminate \(x\) by adding the two equations.

First, add equation (3) and equation (4):

[tex]\[ (x + 4y) + (-x + 2y) = 8 + 10 \][/tex]

This simplifies to:

[tex]\[ (4y + 2y) = 18 \][/tex]

[tex]\[ 6y = 18 \][/tex]

Divide by 6:

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

Now that we have \(y = 3\), substitute \(y\) back into equation (3) to solve for \(x\):

[tex]\[ x + 4(3) = 8 \][/tex]

[tex]\[ x + 12 = 8 \][/tex]

Subtract 12 from both sides:

[tex]\[ x = 8 - 12 \][/tex]

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

Thus, the solution to the system of equations is \( (x, y) = (-4, 3) \).

Given that we have found a unique solution, the system of equations has exactly one solution.