How do you solve by substitution $x + y = 11$ $x + y = 11$ and $x - y = 1$ $x - y = 1$ ?

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Answer 1 · 2015-06-04T04:06:58

The first step would be to write $x - y = 1 \to x = 1 + y$ $x-y=1 rarr x=1+y$
and then to replace $x$ $x$ by $1 + y$ $1+y$ in $x + y = 11$ $x+y=11$

thus $1 + y + y = 11 \to 2 y + 1 = 11 \to 2 y = 10 \to y = 5$ $1+y+y=11 rarr 2y+1=11 rarr 2y=10 rarr y=5$

Now back to $x - y = 1$ $x-y=1$ with $y = 5$ $y=5$

$x - 5 = 1 \to x = 6$ $x-5=1 rarr x=6$

How do you solve by substitution x + y = 11x+y=11 x + y = 11 and x - y = 1x−y=1x - y = 1?