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Phoenix [80]
4 years ago
9

A)Simplify the expression and explain each step. 12 + 3(2y - 3) (B)Factor the expression completely. 18b - 12 (As you solve thes

e problems do it with numbers intead of the word form like 1+1=2 instead of one plus one equals two, please and thank you. I will give brainlyist.
Mathematics
1 answer:
Anika [276]4 years ago
3 0

Answer:

  • A) 6y +3
  • B) 6(3b -2)

Step-by-step explanation:

A) Use the distributive property to eliminate parentheses. Then combine like terms. (The only "like terms" are the constants.)

... = 12 +3·2y +3·(-3) . . . use the distributive property to multiply each term in parentheses by the factor 3 outside those parentheses

... = 12 +6y -9 . . . . . . . . simplify

... = 6y + (12 -9) . . . . . . group like terms together

... = 6y + 3 . . . . . . . . . . simplify

___

B) Look for factors of each term that are also found in the other term.

... 18b has factors 3×6×b

... 12 has factors 2×6

The only common factor is 6, so we factor that out using the distributive property.

... 18b -12 = 6(3b -2)

_____

<em>Comment on factoring</em>

For factoring problems, it helps immensely if you know your times tables and some of the rules for divisibility. (Even numbers are divisible by 2, numbers ending in 0 or 5 are divisible by 5, numbers whose sum of digits is divisible by 3 are divisible by 3, for example.)

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4 years ago
Please help me with this
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Answer:

Option c

Step-by-step explanation:

A system of equations are given to us. And we need to solve them . The given system is

\begin{cases} y = 2x - 3\dots 1 \\ y = x^2 - 3\dots 2 \end{cases}

We numbered the equations here . Now put the value of equation 1 in equation 2 that is substituting y = 2x - 3 in eq. 2 .

\implies y = x^2 - 3 \\\\\implies 2x - 3 = x^2 - 3 \\\\\implies x^2 - 2x = 0 \\\\\implies x(x-2) = 0 \\\\\implies\red{ x = 0 , 2 }

We got two values of x as 0 & 2 . Alternatively substituting these values we have ,

\implies y = 2 x - 3 \\\\\implies y = 2(0)-3 \qquad or \qquad y = 2(2)-3 \\\\\implies y = 0-3 \qquad or \qquad 4 - 3 \\\\\implies \red{ y = -3 , 1 }

Thefore the required answer is ,

\red{Option\:c} \begin{cases} (0,-3) \\ (2,1) \end{cases}

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3 years ago
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Step-by-step explanation:

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Answer:  3\dfrac{2}{6}\ h

Step-by-step explanation:

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