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Nesterboy [21]
3 years ago
8

What is the solution to the inequality |x-4|< 3? –7 < x < –1 1 < x < 7 x < –7 or x < –1 x > 1 or x <

7
Mathematics
1 answer:
balandron [24]3 years ago
5 0

|x-4| < 3\iff x-4 < 3\ \wedge\ x-4 > -3\ \ \ \ |+4\\\\x < 7\ \wedge\ x > 1\\\\Answer:\ \boxed{1 < x < 7}

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Answer:

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Step-by-step explanation:

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We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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Answer:

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Step-by-step explanation:

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