Question

solve the system of equations. if the system has no solution, say that it is inconsistent.
\\( \

$$\begin{cases} 12x - 3y = 9 \\\\ 10x + 3y = 29 \\end{cases}$$

\\)

select the correct choice below and, if necessary, fill in any answer boxes within your choice.

a. the solution is \\( x = \\) \\( \boxed{} \\) and \\( y = \\) \\( \boxed{} \\) (type integers or simplified fractions.)

b. there are infinitely many solutions. using ordered pairs, they can be expressed as \\( \\{ (x, y) \mid x = \\) \\( \boxed{} \\), \\( y \\) any real number \\}
(simplify your answer. type an expression using \\( y \\) as the variable as needed.)

c. the system is inconsistent.

Explanation:

Step1: Add the two equations

We have the system of equations:

$$
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$$

Adding the two equations together to eliminate \(y\):
\((12x - 3y)+(10x + 3y)=9 + 29\)
Simplifying the left - hand side: \(12x+10x-3y + 3y=22x\)
Simplifying the right - hand side: \(38\)
So we get the equation \(22x=38\)

Step2: Solve for \(x\)

Divide both sides of the equation \(22x = 38\) by \(22\):
\(x=\frac{38}{22}=\frac{19}{11}\) Wait, no, let's recalculate. Wait, \(12x+10x = 22x\), \(9 + 29=38\), so \(22x=38\), then \(x=\frac{38}{22}=\frac{19}{11}\)? Wait, no, maybe I made a mistake. Wait, \(12x+10x = 22x\), \( - 3y+3y = 0\), \(9 + 29 = 38\). So \(22x=38\), \(x=\frac{38}{22}=\frac{19}{11}\approx1.727\). Wait, no, wait the original equations: \(12x-3y = 9\) and \(10x + 3y=29\). Let's add them: \(12x+10x-3y + 3y=9 + 29\), so \(22x=38\), \(x=\frac{38}{22}=\frac{19}{11}\). Now substitute \(x = \frac{19}{11}\) into the first equation \(12x-3y=9\)

Step3: Substitute \(x\) into the first equation to solve for \(y\)

Substitute \(x=\frac{19}{11}\) into \(12x-3y = 9\):
\(12\times\frac{19}{11}-3y=9\)
\(\frac{228}{11}-3y = 9\)
Subtract \(\frac{228}{11}\) from both sides:
\(-3y=9-\frac{228}{11}\)
\(9=\frac{99}{11}\), so \(9-\frac{228}{11}=\frac{99 - 228}{11}=\frac{- 129}{11}\)
Then \(-3y=\frac{-129}{11}\)
Divide both sides by \(- 3\):
\(y=\frac{-129}{11}\div(-3)=\frac{-129}{11}\times(-\frac{1}{3})=\frac{43}{11}\)

Wait, no, maybe I made a mistake in addition. Wait, \(12x-3y = 9\) and \(10x + 3y=29\). Let's add them: \(12x + 10x=22x\), \(-3y+3y = 0\), \(9 + 29 = 38\). So \(22x=38\), \(x=\frac{38}{22}=\frac{19}{11}\). Then plug \(x=\frac{19}{11}\) into \(10x + 3y=29\):
\(10\times\frac{19}{11}+3y=29\)
\(\frac{190}{11}+3y=29\)
\(3y=29-\frac{190}{11}\)
\(29=\frac{319}{11}\), so \(29-\frac{190}{11}=\frac{319 - 190}{11}=\frac{129}{11}\)
\(3y=\frac{129}{11}\)
\(y=\frac{129}{11}\div3=\frac{129}{11}\times\frac{1}{3}=\frac{43}{11}\)

Wait, but let's check the addition again. Wait, \(12x-3y = 9\) and \(10x + 3y=29\). Adding gives \(22x=38\), so \(x=\frac{38}{22}=\frac{19}{11}\approx1.727\), and \(y=\frac{43}{11}\approx3.909\). But maybe I made a mistake in the problem reading. Wait, the first equation is \(12x-3y = 9\), the second is \(10x + 3y=29\). Let's add them: \(22x=38\), \(x=\frac{19}{11}\), then from \(12x-3y = 9\), \(12\times\frac{19}{11}= \frac{228}{11}\), \(\frac{228}{11}-9=\frac{228 - 99}{11}=\frac{129}{11}\), then \(-3y=\frac{129}{11}\)? No, wait \(12x-3y = 9\) can be rewritten as \(3y=12x - 9\), so \(y = 4x-3\). Substitute \(y = 4x - 3\) into the second equation \(10x+3y=29\):
\(10x+3(4x - 3)=29\)
\(10x+12x-9 = 29\)
\(22x=29 + 9=38\)
\(x=\frac{38}{22}=\frac{19}{11}\)
Then \(y=4\times\frac{19}{11}-3=\frac{76}{11}-\frac{33}{11}=\frac{43}{11}\)

Wait, but maybe the equations were written incorrectly? Wait, if the first equation was \(12x-3y = 9\) and the second was \(10x + 3y=29\), the solution is \(x=\frac{19}{11}\), \(y=\frac{43}{11}\). But let's check with the addition method again.

Wait, another way: Let's simplify the first equation. Divide \(12x-3y = 9\) by \(3\), we get \(4x-y = 3\), so \(y=4x - 3\). Substitute into the second equation \(10x+3y=29\):

\(10x+3(4x - 3)=29\)

\(10x + 12x-9=29\)

\(22x=38\)

\(x=\frac{38}{22}=\frac{19}{11}\)

\(y=4\times\frac{19}{11}-3=\frac{76 - 33}{11}=\frac{43}{11}\)

Answer:

The solution is \(x = \frac{19}{11}\) and \(y=\frac{43}{11}\)