Question

which system of equations has the same solution as the system below?
$5x + 2y = 33$
$x + 6y = 1$
answer
$5x - 6y = -99$
$x + 6y = 1$
$-15x + 2y = -99$
$x + 6y = 1$
$-15x - 6y = -99$
$x + 6y = 1$
$-15x - 6y = 33$
$x + 6y = 1$
submit answer

Explanation:

Step1: Recall the property of systems of equations

A system of equations has the same solution if we perform a valid row operation (multiplying an equation by a non - zero scalar) on one of the equations. Let's analyze the original system \(

$$\begin{cases}5x + 2y=33\\x + 6y = 1\end{cases}$$

\)

Step2: Analyze the first equation transformation

We want to see which option has a transformed first equation. Let's multiply the first equation \(5x+2y = 33\) by \(- 3\). Using the property \(a(bx+cy)=abx + acy\) (where \(a=-3\), \(b = 5\), \(c = 2\), and the right - hand side \(a\times d\) where \(d = 33\))

Multiplying \(5x+2y=33\) by \(-3\):
\(-3\times(5x + 2y)=-3\times33\)
\(-15x-6y=-99\)? No, wait, let's do it correctly. Wait, if we want to eliminate \(x\) or \(y\), but actually, let's check each option.

Wait, let's multiply the first equation \(5x + 2y=33\) by \(- 3\):

\(-3\times(5x)+(-3)\times(2y)=-3\times33\)

\(-15x-6y=-99\)? No, that's not right. Wait, maybe we made a mistake. Wait, let's check the second option. Wait, no, let's take the original first equation \(5x + 2y=33\). Let's multiply it by \(-3\):

\(-15x-6y=-99\)? No, \(5x\times(-3)=-15x\), \(2y\times(-3)=-6y\), \(33\times(-3)=-99\). Wait, but in the third option, the first equation is \(-15x - 6y=-99\) and the second equation is \(x + 6y=1\). Wait, no, let's check the second equation of the original system is \(x + 6y=1\) (remains the same in all options). Let's check the first equation transformation.

Wait, original first equation: \(5x+2y = 33\). Let's multiply it by \(-3\):

\(-15x-6y=-99\). Wait, but let's check the third option: \(

$$\begin{cases}-15x - 6y=-99\\x + 6y=1\end{cases}$$

\)

If we add the two equations in the third option: \((-15x-6y)+(x + 6y)=-99 + 1\), \(-14x=-98\), \(x = 7\). Let's check with the original system.

From the original system, from the second equation \(x=1 - 6y\). Substitute into the first equation: \(5(1 - 6y)+2y=33\), \(5-30y+2y=33\), \(5-28y=33\), \(-28y=28\), \(y=-1\), then \(x=1-6\times(-1)=7\).

Now check the third option: \(-15x-6y=-99\) and \(x + 6y=1\). Add the two equations: \(-15x-6y+x + 6y=-99 + 1\), \(-14x=-98\), \(x = 7\). Then substitute \(x = 7\) into \(x + 6y=1\), \(7+6y=1\), \(6y=-6\), \(y=-1\). Which is the same solution as the original system.

Wait, but let's check the other options.

First option: First equation \(5x-6y=-99\), second equation \(x + 6y=1\). Add them: \(6x=-98\), \(x=-\frac{49}{3}\), which is different from the original solution (\(x = 7\)), so first option is wrong.

Second option: First equation \(-15x+2y=-99\), second equation \(x + 6y=1\). Let's solve: from second equation \(x=1 - 6y\). Substitute into first equation: \(-15(1 - 6y)+2y=-99\), \(-15 + 90y+2y=-99\), \(92y=-84\), \(y=-\frac{21}{23}\), which is different from \(y=-1\), so second option is wrong.

Fourth option: First equation \(-15x-6y=33\), second equation \(x + 6y=1\). Add them: \(-14x=34\), \(x=-\frac{17}{7}\), different from \(x = 7\), so fourth option is wrong.

Third option: First equation \(-15x-6y=-99\), second equation \(x + 6y=1\). As we saw, adding them gives \(-14x=-98\), \(x = 7\), and then \(y=-1\), which is the same as the original system.

Answer:

\(\boldsymbol{

$$\begin{cases}-15x - 6y=-99\\x + 6y=1\end{cases}$$

}\) (the third option: \(-15x - 6y=-99\) and \(x + 6y=1\))