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Select all the systems of equations that have no solutions. 1)5x – 2y = 3 10x - 4y = 6 2)3x + 7y = 42 6x + 14y = 503)y=5 + 2xy=5x + 2 4)y=4x-14y=16x-45)y=-3x+5y=-3x+4

User Ibaralf
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1 Answer

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The first step to determine if an equation system has or not a solution is to simplify the equations to reduce them to their simpliest form and rearragne the terms so that the variables are in the same order in both equations.

Then you have to compare them.

System 1


\begin{gathered} 5x-2y=3 \\ 10x-4y=6 \end{gathered}

The first equation is reduced but the second one is not. Divide the second equation by 2 to simplify it:


\begin{gathered} (10x)/(2)-(4y)/(2)=(6)/(2) \\ 5x-2y=3 \end{gathered}

Now you can compare them


\begin{gathered} 5x-2y=3 \\ 5x-2y=3 \end{gathered}
\begin{gathered} 5x-2y=3 \\ 5x=3+2y \\ x=(3)/(5)+(2)/(5)y \end{gathered}

Replace


\begin{gathered} 5((3)/(5)+(2)/(5)y)-2y=3 \\ 3+2y-2y=3 \\ 3=3 \end{gathered}

The result is a true statement, this system has infinite solutions.

System 2


\begin{gathered} 3x+7y=42 \\ 6x+14y=50 \end{gathered}

First step is to reduce the second equation, to do so, divide it by 2


\begin{gathered} (6x)/(2)+(14y)/(2)=(50)/(2) \\ 3x+7y=25 \end{gathered}

Solve the first equation for x:


\begin{gathered} 3x+7y=42 \\ 3x=42-7y \\ (3x)/(3)=(42)/(3)-(7)/(3)y \\ x=14-(7)/(3)y \end{gathered}

Replace it in the econd equation:


\begin{gathered} 3(14-(7)/(3)y)+7y=25 \\ 42-7y+7y=25 \\ 42=25 \end{gathered}

This sistem leads to a false statement and therefore has no result.

System 3


\begin{gathered} y=5+2x \\ y=5x+2 \end{gathered}

THe equations are already in their most reduced form, replace the first one in the second one and solve for x:


\begin{gathered} 5+2x=5x+2 \\ 5-2=5x-2x \\ 3=3x \\ x=1 \end{gathered}

Now replace the value of x and solve fot y in the first one


\begin{gathered} y=5+2\cdot1 \\ y=7 \end{gathered}

This equations system has one possible solution for x=1 and y=7

System 4


\begin{gathered} y=4x-1 \\ 4y=16x-4 \end{gathered}

First step is to divide the second equation by 4 to simplify it


\begin{gathered} (4y)/(4)=(16x)/(4)-(4)/(4) \\ y=4x-1 \end{gathered}

Both equations are equal, replace the first formula in the second one and solve for x:


\begin{gathered} 4x-1=4x-1 \\ 4x-4x=1-1 \\ 0=0 \end{gathered}

This system has infinite solutions.

System 5


\begin{gathered} y=-3x+5 \\ y=-3x+4 \end{gathered}

Both equations are in their most reduced form, replece the first one in the second one and solve for x:


\begin{gathered} -3x+5=-3x+4 \\ -3x+3x=4+5 \\ 0=9 \end{gathered}

This equation system leads to a false statement, this means that it has no solution.

So, the equations systems that have no solution are the second and fifth systems

User Ghita
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