The answer is: [C]: " {x | x > 5} " .
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Explanation:
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Given: " 4(x − 3) − 2(x − 1) > 0 " ;
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Note the "distributive property of multiplication" :
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a(b+c) = ab + ac ;
a(b−c) = ab − ac ;
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So, given:
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→ 4(x − 3) − 2(x − 1) > 0 ;
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Let us simplify; and rewrite:
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Start with:
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→ -2 (x − 1) = (-2*x) − (-2 *1) = -2x − (-2) = -2x + 2 ;
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Now, continue with:
→ 4(x − 3) = (4*x) − (4*3) = 4x − 12 ;
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So, given the original problem:
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→ 4(x − 3) − 2(x − 1) > 0 ;
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Rewrite;
Replacing: "4(x − 3)" ; with: "4x − 12" ;
and replacing "− 2(x − 1)" ; with: " -2x + 2" ;
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as follows: → " 4x − 12 − 2x + 2 " > 0 ;
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On the "left-hand side", combine the "like terms", and simplify ;
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+4x −2x = +2x ; −12 +2 = -10 ; and rewrite:
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→ 2x − 10 > 0 ; Add "10" to EACH SIDE of the inequality;
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→ 2x − 10 + 10 > 0 + 10 ;
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to get: → 2x > 10 ;
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→ Now, divide EACH SIDE of the inequality by "2";
to isolate "x" on one side of the inequality; & to "solve"/"simply" for "x" ;
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→ 2x / 2 > 10 / 2 ;
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→ x > 5 ; which is:
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→ Answer choice: [C]: " {x | x > 5} " .
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