# Solve the system of equations: 3x - y = -2, 2x^2 - y = 0 Question
Equations Solve the system of equations: $$\displaystyle{3}{x}-{y}=-{2},{2}{x}^{{2}}-{y}={0}$$ 2020-12-23
Step 1
Given: The system of equations
3x−y=−2
$$\displaystyle{2}{x}^{{2}}−{y}={0}$$
To find: The solution of given system of equation.
Step 2
Explanation:
Let
3x−y=−2 ...(i)
$$\displaystyle{2}{x}^{{2}}−{y}={0}$$ ...(ii),
From equation (i),
3x−y=−2,
$$\displaystyle\Rightarrow{y}={3}{x}+{2}$$
Substituting y=3x+2 in equation (ii),
$$\displaystyle{2}{x}^{{2}}-{y}={0}$$
$$\displaystyle\Rightarrow{2}{x}^{{2}}-{\left({3}{x}+{2}\right)}={0}$$
$$\displaystyle\Rightarrow{2}{x}^{{2}}-{3}{x}-{2}={0}$$
$$\displaystyle\Rightarrow{2}{x}^{{2}}-{4}{x}+{x}-{2}={0}$$
$$\displaystyle\Rightarrow{\left({x}-{2}\right)}{\left({2}{x}+{1}\right)}={0}$$
$$\displaystyle\Rightarrow{x}={2},-\frac{{1}}{{2}}$$
Now,
For $$\displaystyle{x}={2},{y}={3}\times{2}+{2}={8}$$,
For $$\displaystyle{x}=-\frac{{1}}{{2}},{y}={3}\times{\left(-\frac{{1}}{{2}}\right)}+{2}=-\frac{{3}}{{2}}+{2}=\frac{{1}}{{2}}$$
Hence x=2, y=8 and $$\displaystyle{x}=−\frac{{1}}{{2}},{y}=\frac{{1}}{{2}}$$ are solution of given system of equations.
Result: x=2, y=8 and $$\displaystyle{x}=−\frac{{1}}{{2}},{y}=\frac{{1}}{{2}}$$ are solution of given system of equation.

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