Solve for the value of x

Question

Asked: March 30, 20212021-03-30T03:32:27+00:00 2021-03-30T03:32:27+00:00In: Algebra, High School/Senior Secondary School

1. 9x-81=0

2. 5x+2=27

3. 5(x+2)=3(x+10)

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acedstud · Accepted Answer · 2021-03-30T08:21:33+00:00

1.) Given: $9x-81=0$

Adding 81 to both sides, we have $9x-81+81=0+81\Rightarrow9x=81$ .

Dividing both sides by the coefficient of $x$ , which is 9, we have $\dfrac{9x}{9}=\dfrac{81}{9}\Rightarrow x=9$ .

Therefore, $x=9$ .

2.) Given: $5x+2=27$

Subtracting 2 from both sides, we have $5x+2-2=27-2\Rightarrow5x=25$ .

Dividing both sides by the coefficient of $x$ , which is 5, we have $\dfrac{5x}{5}=\dfrac{25}{5}\Rightarrow x=5$ .

Therefore, $x=5$ .

c.) Given: $5(x+2)=3(x+10)$

Using foil/distributive law to expand, we have $5x+10=3x+30$ .

Subtracting $3x$ from both sides gives $5x+10-3x=3x+30-3x\Rightarrow2x+10=30$ .

Subtracting 10 from both sides gives $2x+10-10=30-10\Rightarrow2x=20$ .

Dividing both sides by 2 gives $\dfrac{2x}{2}=\dfrac{20}{2}\Rightarrow x=10$ .

Therefore, $x=10$ .

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