Question

What is the solution of this system of linear equations?

3y = 3 y equals StartFraction 3 over 2 EndFraction x plus 6.x + 6

y – StartFraction one-half EndFraction y minus StartFraction 1 over 4 EndFraction x equals 3.x = 3

(3, 6)
(2, 1)
no solution
infinite number of solutions

Answer 1

To solve the system of linear equations you provided, we need to express each equation in a standard form and then analyze them. Let's first rewrite the equations clearly.

1. The first equation is:
\[
3y = \frac{3}{2}x + 6.
\]
We can rearrange this to:
\[
3y - \frac{3}{2}x - 6 = 0.
\]

2. The second equation is:
\[
y - \frac{1}{2}y - \frac{1}{4}x = 3.
\]
We simplify this:
\[
\frac{1}{2}y - \frac{1}{4}x = 3.
\]
Rearranging gives us:
\[
\frac{1}{2}y = \frac{1}{4}x + 3.
\]
This can be rewritten as:
\[
1y - \frac{1}{2}x - 6 = 0.
\]

Now, we compose a system of equations:
1. \(3y - \frac{3}{2}x - 6 = 0\) (Let's call this equation 1)
2. \(y - \frac{1}{2}x - 6 = 0\) (Let's call this equation 2)

To solve this system, we can manipulate these equations to analyze them further:

From equation 2:
\[
y = \frac{1}{2}x + 6.
\]

Now, we can substitute the value of \(y\) into equation 1:
\[
3\left(\frac{1}{2}x + 6\right) - \frac{3}{2}x - 6 = 0.
\]
Expanding this:
\[
\frac{3}{2}x + 18 - \frac{3}{2}x - 6 = 0,
\]
which simplifies to:
\[
18 - 6 = 0 \rightarrow 12 = 0.
\]
This is a contradiction, indicating that the two equations do not intersect.

Since we arrive at a contradiction, the conclusion is that the system of equations has **no solution**. Therefore, the correct answer is "no solution."