ladymint0w

2022-10-03

Use substitution to solve $$2x+y+z=4,\text{}2x-y-z=1$$ and $$2x-y+z=3$$

Nolan Tyler

Beginner2022-10-04Added 9 answers

The set of equations $$2x+y+z=4,\text{}2x-y-z=1$$ and $$2x-y+z=3$$ has to be solved for x, y and z.

Isolate y from $$2x+y+z=4,\text{}y=4-2x-z$$

Substitute $$\Rightarrow 2x-4+2x+z-z=1\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=5\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{4}\phantom{\rule{0ex}{0ex}}y=4-2\cdot \frac{5}{4}-z\phantom{\rule{0ex}{0ex}}y=1.5-z$$

Substitute $$\Rightarrow 2.5-1.5+z+z=3\phantom{\rule{0ex}{0ex}}\Rightarrow 2z=2\phantom{\rule{0ex}{0ex}}\Rightarrow z=1\phantom{\rule{0ex}{0ex}}y=1.5-z=1.5-1=0.5$$

Isolate y from $$2x+y+z=4,\text{}y=4-2x-z$$

Substitute $$\Rightarrow 2x-4+2x+z-z=1\phantom{\rule{0ex}{0ex}}\Rightarrow 4x=5\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{4}\phantom{\rule{0ex}{0ex}}y=4-2\cdot \frac{5}{4}-z\phantom{\rule{0ex}{0ex}}y=1.5-z$$

Substitute $$\Rightarrow 2.5-1.5+z+z=3\phantom{\rule{0ex}{0ex}}\Rightarrow 2z=2\phantom{\rule{0ex}{0ex}}\Rightarrow z=1\phantom{\rule{0ex}{0ex}}y=1.5-z=1.5-1=0.5$$

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