Respuesta :

SOLVING THE INEQUALITY  [tex]3x-10<-25[/tex] TO MATCH OPTION A

Considering the inequality

[tex]3x-10<-25[/tex]

[tex]3x-10+10<-25+10[/tex]

[tex]3x<-15[/tex]

[tex]\frac{3x}{3}<\frac{-15}{3}[/tex]

[tex]x<-5[/tex]

Thus,

[tex]3x-10<-25\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<-5\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-5\right)\end{bmatrix}[/tex]

Therefore, the inequality [tex]3x-10<-25[/tex] has the solution [tex]x<-5[/tex] which matches the option A.

SOLVING THE INEQUALITY  [tex]-4x+7\ge \:27[/tex]  TO MATCH OPTION B

Considering the inequality

[tex]-4x+7\ge \:27[/tex]

[tex]-4x+7-7\ge \:27-7[/tex]

[tex]-4x\ge \:20[/tex]

[tex]\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}[/tex]

[tex]\left(-4x\right)\left(-1\right)\le \:20\left(-1\right)[/tex]

[tex]4x\le \:-20[/tex]

[tex]\frac{4x}{4}\le \frac{-20}{4}[/tex]

[tex]x\le \:-5[/tex]

Thus,

[tex]-4x+7\ge \:27\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-5\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-5]\end{bmatrix}[/tex]

Therefore, the inequality  [tex]-4x+7\ge \:27[/tex] has the solution [tex]x\le \:-5[/tex] which matches the option B.

SOLVING THE INEQUALITY  [tex]\frac{x}{1}-9<-10[/tex]  TO MATCH OPTION C

Considering the inequality

[tex]\frac{x}{1}-9<-10[/tex]

[tex]\mathrm{Apply\:rule}\:\frac{a}{1}=a[/tex]

[tex]x-9<-10[/tex]     ∵  [tex]\frac{x}{1}=x[/tex]

[tex]x-9<-10[/tex]

[tex]x-9+9<-10+9[/tex]

[tex]x<-1[/tex]

[tex]\frac{x}{1}-9<-10\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x<-1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-1\right)\end{bmatrix}[/tex]

Therefore, the inequality [tex]\frac{x}{1}-9<-10[/tex] has the solution [tex]x<-1[/tex] which matches the option C.

SOLVING THE INEQUALITY [tex]5\left(x-2\right)>-15[/tex] TO MATCH OPTION D

Considering the inequality

[tex]5\left(x-2\right)>-15[/tex]

[tex]\mathrm{Divide\:both\:sides\:by\:}5[/tex]

[tex]\frac{5\left(x-2\right)}{5}>\frac{-15}{5}[/tex]

[tex]x-2>-3[/tex]

[tex]x-2+2>-3+2[/tex]

[tex]x>-1[/tex]

Thus,

[tex]5\left(x-2\right)>-15\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x>-1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-1,\:\infty \:\right)\end{bmatrix}[/tex]

Therefore, the inequality [tex]5\left(x-2\right)>-15[/tex] has the solution [tex]x>-1[/tex] which matches the option D.

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