\left\{ \begin{array} { l } { a = 3 } \\ { a + x = 1 } \end{array} \right.
Solve for a, x
x=-2
a=3
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\left\{ \begin{array} { l } { a = 3 } \\ { a + x = 1 } \end{array} \right.
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3+x=1
Consider the second equation. Insert the known values of variables into the equation.
x=1-3
Subtract 3 from both sides.
x=-2
Subtract 3 from 1 to get -2.
a=3 x=-2
The system is now solved.
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