Solve the given two equations and select the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Solve the first equation $x^{2}-6x=7$.$x^{2}-6x-7=0$$x^{2}-7x+x-7=0$$x(x-7)+1(x-7)=0$$(x+1)(x-7)=0$So,$x = -1$ or $x = 7$.Step $2$: Solv

Vedclass · Accepted Answer

If $x > y$

Vedclass · Answer

(B) Step $1$: Solve the first equation $x^{2}-6x=7$.$x^{2}-6x-7=0$$x^{2}-7x+x-7=0$$x(x-7)+1(x-7)=0$$(x+1)(x-7)=0$So,$x = -1$ or $x = 7$.Step $2$: Solve the second equation $2y^{2}+13y+15=0$.$2y^{2}+10y+3y+15=0$$2y(y+5)+3(y+5)=0$$(2y+3)(y+5)=0$So,$y = -1.5$ or $y = -5$.Step $3$: Compare the values.Possible values for $x$ are $\{-1, 7\}$.Possible values for $y$ are $\{-1.5, -5\}$.Comparing the values: Since $-1 > -1.5$,$-1 > -5$,$7 > -1.5$,and $7 > -5$,it is clear that $x > y$ in all cases.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}-6x=7$
$II.$ $2y^{2}+13y+15=0$

Similar Questions

If $x-\frac{1}{x}=\sqrt{21},$ then the value of $\left(x^{2}+\frac{1}{x^{2}}\right)\left(x+\frac{1}{x}\right)$ will be

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0, a \in R$ and $\alpha < 1 < \beta$,then:

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 49$
$II.$ $y^{2} + 15y + 56 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{9}{\sqrt{x}} + \frac{19}{\sqrt{x}} = \sqrt{x}$
$II.$ $y^{5} - \frac{(2 \times 14)^{11/2}}{\sqrt{y}} = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt{172}$
$II.$ $y^{2} - 29y + 210 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $x^{2}-6x=7$$II.$ $2y^{2}+13y+15=0$