The values of $y$ which will satisfy the equations
$2 x^{2} + 6 x + 5 y + 1 = 0, 2 x + y + 3 = 0$ may be found by solving.

y^{2} + 14 y - 7 = 0

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y^{2} + 8 y + 1 = 0

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y^{2} + 10 y - 7 = 0

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y^{2} + y - 12 = 0

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None of these

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Solution

The correct option is E $y^{2} + 10 y - 7 = 0$
Solve the second equation for $x$ , substitute into the first equation and simplify:
$x = - \frac{y + 3}{2}, 2 {(- \frac{y + 3}{2})}^{2} + 6 (- \frac{y + 3}{2}) + 5 y + 1 = 0$ ;
$∴ y^{2} + 10 y - 7 = 0$ .

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Similar questions

Q. The value of

y

which will satisfy the equations

2 x^{2} + 6 x + 5 y + 1 = 0

2 x + y + 3 = 0

may be found by solving which one the following equations?

Q. Find the equation of the circle which cuts the following circles orthogonally.
(i)

x^{2} + y^{2} + 4 x - 7 = 0

2 x^{2} + 2 y^{2} + 3 x + 5 y - 9 = 0

x^{2} + y^{2} + y = 0

Q. Find the maximum value of y which satisfy the inequations

y^{2} + 5 y + 4 \leq 0

and

y^{2} - 2 y - 15 \geq 0

Find the equation of the circle which cuts the following circles orthogonally.
$x^{2} + y^{2} + 2 x + 4 y + 1 = 0,$
$2 (x^{2} + y^{2}) + 6 x + 8 y - 3 = 0,$
$x^{2} + y^{2} - 2 x + 6 y - 3 = 0$

Q. The centre of the circles

x^{2} - y^{2} - 6 x - 8 y - 7 = 0

and

x^{2} - y^{2} - 4 x - 10 y - 3 = 0

are the ends of the diameter of the circle:

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The values of y which will satisfy the equations2x2+6x+5y+1=0,2x+y+3=0 may be found by solving.

The correct option is E y2+10y−7=0Solve the second equation for x, substitute into the first equation and simplify:x=−y+32,2(−y+32)2+6(−y+32)+5y+1=0;∴y2+10y−7=0.

The values of $y$ which will satisfy the equations
$2 x^{2} + 6 x + 5 y + 1 = 0, 2 x + y + 3 = 0$ may be found by solving.

The correct option is E $y^{2} + 10 y - 7 = 0$
Solve the second equation for $x$ , substitute into the first equation and simplify:
$x = - \frac{y + 3}{2}, 2 {(- \frac{y + 3}{2})}^{2} + 6 (- \frac{y + 3}{2}) + 5 y + 1 = 0$ ;
$∴ y^{2} + 10 y - 7 = 0$ .