Solve the following system of equations:
Solve the following system of equations:
\( \begin{cases}
x+y=8 \\
x=5-y
\end{cases} \)
\( x+y=8 \)
\( x-y=6 \)
\( 3x-y=5 \)
\( 5x+2y=12 \)
\( 6x+4y=18 \)
\( -2x+3y=20 \)
\( 6x+y=12 \)
\( 3y+2x=20 \)
Solve the following system of equations:
Note that in the current system of equations, one of the variables is isolated alone on the left side of the equation:
Therefore, we can use the substitution method and substitute the entire expression that x equals in the second equation in place of x in the first equation (marked with an underline in both equations above) and thus we'll get one equation with one variable:
where we highlight the equation in which the variable we substituted is isolated in order to return to it later after we find the value of y from solving the equation we got, and this is to find using it the value of x corresponding to that y value we found, therefore we highlighted this equation above.
From here - we'll continue and solve the single-variable equation we got,
First- we'll combine like terms on the left side of the resulting equation:
We'll stop here and notice that y cancelled out in the current equation and we got a false statement, this is because clearly:
meaning-
We got a false statement regardless of the variables' values,
We can conclude from here that the system of equations has no solution, because no matter which values we substitute for the variables - we won't get a true statement in both equations together.
Therefore the correct answer is answer D.
There is no solution.
\( -x+y=14 \)
\( 5x+2y=7 \)
\( 4x+3y=-11 \)
\( 3x-2y=-4 \)
\( 4x-8y=16 \)
\( -x-2y=24 \)
\( 6x-2y=24 \)
\( x+5y=4 \)
\( x-y=8 \)
\( 2x-2y=16 \)
Infinite solutions
\( x-y=8 \)
\( 3x+2y=24 \)
\( 5y+3x=15 \)
\( -2y-4x=-34 \)