Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Maya was asked whether the following equation is an identity:

(2x+3)(x+1)=2(x+1)^(2)+(x+1)
She performed the following steps:

(2x+3)(x+1)

longrightarrow^(" Step "1)=2x^(2)+2x+3x+3

longrightarrow^(" Step "2)=2x^(2)+4x+2+x+1

harr^(" Step "3)=2(x^(2)+2x+1)+(x+1)

longrightarrow^(" Step "4)=2(x+1)^(2)+(x+1)
For this reason, Maya stated that the equation is a true identity.
Is Maya correct? If not, in which step did she make a mistake?
Choose 1 answer:
(A) Maya is correct.
(B) Maya is incorrect. She made a mistake in step 1.
(C) Maya is incorrect. She made a mistake in step 2 .
(D) Maya is incorrect. She made a mistake in step 4.

Maya was asked whether the following equation is an identity:\newline(2x+3)(x+1)=2(x+1)2+(x+1) (2 x+3)(x+1)=2(x+1)^{2}+(x+1) \newlineShe performed the following steps:\newline(2x+3)(x+1) (2 x+3)(x+1) \newline Step 1=2x2+2x+3x+3 \stackrel{\text { Step } 1}{\hookrightarrow}=2 x^{2}+2 x+3 x+3 \newline Step 2=2x2+4x+2+x+1 \stackrel{\text { Step } 2}{\hookrightarrow}=2 x^{2}+4 x+2+x+1 \newline Step 3=2(x2+2x+1)+(x+1) \stackrel{\text { Step } 3}{\hookrightarrow}=2\left(x^{2}+2 x+1\right)+(x+1) \newline Step 4=2(x+1)2+(x+1) \stackrel{\text { Step } 4}{\hookrightarrow}=2(x+1)^{2}+(x+1) \newlineFor this reason, Maya stated that the equation is a true identity.\newlineIs Maya correct? If not, in which step did she make a mistake?\newlineChoose 11 answer:\newline(A) Maya is correct.\newline(B) Maya is incorrect. She made a mistake in step 11.\newline(C) Maya is incorrect. She made a mistake in step 22 .\newline(D) Maya is incorrect. She made a mistake in step 44.

Full solution

Q. Maya was asked whether the following equation is an identity:\newline(2x+3)(x+1)=2(x+1)2+(x+1) (2 x+3)(x+1)=2(x+1)^{2}+(x+1) \newlineShe performed the following steps:\newline(2x+3)(x+1) (2 x+3)(x+1) \newline Step 1=2x2+2x+3x+3 \stackrel{\text { Step } 1}{\hookrightarrow}=2 x^{2}+2 x+3 x+3 \newline Step 2=2x2+4x+2+x+1 \stackrel{\text { Step } 2}{\hookrightarrow}=2 x^{2}+4 x+2+x+1 \newline Step 3=2(x2+2x+1)+(x+1) \stackrel{\text { Step } 3}{\hookrightarrow}=2\left(x^{2}+2 x+1\right)+(x+1) \newline Step 4=2(x+1)2+(x+1) \stackrel{\text { Step } 4}{\hookrightarrow}=2(x+1)^{2}+(x+1) \newlineFor this reason, Maya stated that the equation is a true identity.\newlineIs Maya correct? If not, in which step did she make a mistake?\newlineChoose 11 answer:\newline(A) Maya is correct.\newline(B) Maya is incorrect. She made a mistake in step 11.\newline(C) Maya is incorrect. She made a mistake in step 22 .\newline(D) Maya is incorrect. She made a mistake in step 44.
  1. Combine Terms: Combine like terms.\newline2x2+2x+3x+3=2x2+5x+32x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3
  2. Rewrite Equation: Rewrite the right-hand side of the equation.\newline2(x+1)2+(x+1)=2(x2+2x+1)+x+12(x+1)^2 + (x+1) = 2(x^2 + 2x + 1) + x + 1
  3. Expand Right Side: Expand the right-hand side.\newline2(x2+2x+1)+x+1=2x2+4x+2+x+12(x^2 + 2x + 1) + x + 1 = 2x^2 + 4x + 2 + x + 1
  4. Combine Like Terms: Combine like terms on the right-hand side.\newline2x2+4x+2+x+1=2x2+5x+32x^2 + 4x + 2 + x + 1 = 2x^2 + 5x + 3
  5. Compare Sides: Compare the left-hand side and the right-hand side. Since both sides are equal, Maya's equation is an identity.

More problems from Transformations of quadratic functions