defazajx
2020-12-22
Answered

Solve the system of equations: $3x-y=-2,2{x}^{2}-y=0$

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odgovoreh

Answered 2020-12-23
Author has **107** answers

Step 1

Given: The system of equations

3x−y=−2

$2{x}^{2}-y=0$

To find: The solution of given system of equation.

Step 2

Explanation:

Let

3x−y=−2 ...(i)

$2{x}^{2}-y=0$ ...(ii),

From equation (i),

3x−y=−2,

$\Rightarrow y=3x+2$

Substituting y=3x+2 in equation (ii),

$2{x}^{2}-y=0$

$\Rightarrow 2{x}^{2}-(3x+2)=0$

$\Rightarrow 2{x}^{2}-3x-2=0$

$\Rightarrow 2{x}^{2}-4x+x-2=0$

$\Rightarrow (x-2)(2x+1)=0$

$\Rightarrow x=2,-\frac{1}{2}$

Now,

For$x=2,y=3\times 2+2=8$ ,

For$x=-\frac{1}{2},y=3\times (-\frac{1}{2})+2=-\frac{3}{2}+2=\frac{1}{2}$

Hence x=2, y=8 and$x=-\frac{1}{2},y=\frac{1}{2}$ are solution of given system of equations.

Result: x=2, y=8 and$x=-\frac{1}{2},y=\frac{1}{2}$ are solution of given system of equation.

Given: The system of equations

3x−y=−2

To find: The solution of given system of equation.

Step 2

Explanation:

Let

3x−y=−2 ...(i)

From equation (i),

3x−y=−2,

Substituting y=3x+2 in equation (ii),

Now,

For

For

Hence x=2, y=8 and

Result: x=2, y=8 and

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