which expressions are equivalent??
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Given:
The given expression is [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex]
We need to determine the equivalent expression.
Option A: -1
Solving the expression, we get;
[tex]\frac{3}{2} x+14-\frac{3}{2} x+15[/tex]
Simplifying, we get;
[tex]14+15=29[/tex]
Thus, the expression [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex] is not equivalent to -1.
Hence, Option A is not the correct answer.
Option B: 29
Solving the expression, we get;
[tex]\frac{3}{2} x+14-\frac{3}{2} x+15[/tex]
Simplifying, we get;
[tex]14+15=29[/tex]
Thus, the expression [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex] is equivalent to 29.
Hence, Option B is the correct answer.
Option C: [tex]2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right][/tex]
Let us rewrite the given expression.
[tex]2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right][/tex]
Thus, the expression [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex] is equivalent to [tex]2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right][/tex]
Thus, Option C is the correct answer.
Option D: [tex]2\left(\frac{3}{4} x\right)+2(7)+3\left(\frac{1}{2} x\right)+3(-5)[/tex]
Rewriting the expression, we get;
[tex]2\left(\frac{3}{4} x+7\right)+(-3)\left[\frac{1}{2} x+(-5)\right][/tex]
Hence, the expression [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex] is equivalent not to [tex]2\left(\frac{3}{4} x\right)+2(7)+3\left(\frac{1}{2} x\right)+3(-5)[/tex]
Thus, Option D is not the correct answer.
Option E: [tex]2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)[/tex]
Multiplying the terms within the bracket, we get;
[tex]2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)[/tex]
Hence, the expression [tex]2\left(\frac{3}{4} x+7\right)-3\left(\frac{1}{2} x-5\right)[/tex] is equivalent to [tex]2\left(\frac{3}{4} x\right)+2(7)+(-3)\left(\frac{1}{2} x\right)+(-3)(-5)[/tex]
Thus, Option E is the correct answer.