Answer:
(2;5)
Step-by-step explanation:
Elimination:
Let's first multiply the first by 2 to clear fractions and write them both in standard form:
[tex]\left\{ {x-2y=-8} \atop {2x+y=9}} \right. \\[/tex]
Eliminate x: Multiply the first by 2, then subtract the equations:
[tex]2I-II: 2x-4y-2x-y=-16-9\\-5y=-25 \rightarrow y= 5[/tex]
Eliminate y: Multiply the second by 2, then add equations:
[tex]I+2II: x-2y +4x+2y = -8+18\\5x=10 \rightarrow x=2[/tex]
Subtsitution
Solve the first for y (oh, what a coincidence, it's done already!) and replace the value in the second.
[tex]\frac12x +4 = -2x+9 \rightarrow \ x+8=-4x+18\\x+4x = 18-8 \rightarrow 5x=10 \rightarrow x=2[/tex]
Then let's plug this value into a convenient one - let's say, the first - and find y from there:
[tex]y=\frac12\cdot 2 +4\rightarrow y =1+4 \rightarrow y=5[/tex]
If anyone tries saying it's not substitution, bring the system in standard form, as you did for the elimination method, then solve again for y.
