Question

y = 5x - 1
-15x - 3y = 3
how many solutions does this linear system have?
one solution: (0, -1)
one solution: (1, 4)
no solution
infinite number of solutions

Explanation:

Step1: Rewrite the second equation

We have the second equation \(-15x - 3y = 3\). Let's divide the entire equation by \(-3\) to simplify it.
\(\frac{-15x}{-3}-\frac{3y}{-3}=\frac{3}{-3}\)
Which simplifies to \(5x + y=- 1\), or we can rewrite it as \(y=-5x - 1\)? Wait, no, wait. Wait, original first equation is \(y = 5x-1\). Wait, let's do the simplification again.

Wait, \(-15x-3y = 3\), divide both sides by \(-3\):

\(\frac{-15x}{-3}+\frac{-3y}{-3}=\frac{3}{-3}\)

So \(5x + y=-1\), then we can solve for \(y\): \(y=-5x - 1\)? Wait, that can't be. Wait, no, wait, let's substitute the first equation \(y = 5x-1\) into the second equation.

Substitute \(y = 5x - 1\) into \(-15x-3y=3\):

\(-15x-3(5x - 1)=3\)

Step2: Expand and simplify

Expand the left - hand side: \(-15x-15x + 3=3\)

Combine like terms: \(-30x+3 = 3\)

Subtract 3 from both sides: \(-30x=0\)

Then \(x = 0\)

Step3: Find the value of y

Substitute \(x = 0\) into the first equation \(y = 5x-1\), we get \(y=5\times0 - 1=-1\)

Wait, but let's check the slopes of the two lines. The first line \(y = 5x-1\) has a slope \(m_1 = 5\). Let's rewrite the second equation in slope - intercept form (\(y=mx + b\)).

Starting with \(-15x-3y = 3\), we can solve for \(y\):

\(-3y=15x + 3\)

Divide both sides by \(-3\): \(y=-5x - 1\). Wait, no, that's a mistake earlier. Wait, \(-15x-3y = 3\), add \(15x\) to both sides: \(-3y=15x + 3\), then divide by \(-3\): \(y=-5x - 1\). Wait, but the first equation is \(y = 5x-1\). Wait, no, that can't be. Wait, I must have made a mistake in substitution.

Wait, let's do the substitution again. The first equation is \(y = 5x-1\). Substitute into the second equation \(-15x-3y = 3\):

\(-15x-3(5x - 1)=3\)

\(-15x-15x + 3=3\)

\(-30x+3 = 3\)

\(-30x=0\)

\(x = 0\), then \(y=5(0)-1=-1\)

Now, let's check if \((0, - 1)\) satisfies the second equation:

Left - hand side of \(-15x-3y\) when \(x = 0,y=-1\): \(-15(0)-3(-1)=0 + 3=3\), which is equal to the right - hand side.

But wait, let's check the slopes. The first line \(y = 5x-1\) has slope \(m_1 = 5\). The second line, from \(-15x-3y = 3\), we can rewrite it as \(y=-5x - 1\), slope \(m_2=-5\)? Wait, no, that's wrong. Wait, \(-15x-3y = 3\)

\(-3y=15x + 3\)

\(y=-5x - 1\). Wait, but the first equation is \(y = 5x-1\). Wait, that would mean the slopes are different (\(5\) and \(-5\)), so they should intersect at one point. But when we substituted, we got \(x = 0,y=-1\). Wait, but let's check the second equation again. Wait, maybe I made a mistake in rewriting the second equation.

Wait, original second equation: \(-15x-3y = 3\). Let's divide by \(-3\): \(5x + y=-1\), so \(y=-5x - 1\). First equation: \(y = 5x-1\).

Set them equal: \(5x-1=-5x - 1\)

\(5x + 5x=-1 + 1\)

\(10x=0\)

\(x = 0\), then \(y=5(0)-1=-1\). So the solution is \((0,-1)\), which means there is one solution \((0,-1)\)

Wait, but earlier when I thought the slope was \(-5\) for the second line and \(5\) for the first, that's correct. So two lines with different slopes will intersect at exactly one point. And we found that point is \((0,-1)\)

Answer:

one solution: \((0, - 1)\)