Question

1. vic solved the equation ( 0.2k + 2 = 1.4k - 1 ) as shown but made an error in his work.

( 0.2k + 2 = 1.4k - 1 )
( 10(0.2k + 2)=10(1.4k - 1) )
( 2k + 2 = 14k - 1 )
( 2k+(-14k)+2 = 14k+(-14k)-1 )
( -12k + 2=-1 )
( -12k + 2+(-2)=-1+(-2) )
( -12k=-3 )
( -\frac{1}{12}(-12k)=-\frac{1}{12}(-3) )
( k=\frac{1}{4} )
the solution is ( \frac{1}{4} ).
a. describe vics error.
b. find the correct value for ( k ).

1. in the diagram, ( overleftrightarrow{de} parallel overleftrightarrow{fh} ). find the measure of ( angle jge ).

Explanation:

Step1: Analyze Vic's error

When Vic multiplied \(10(0.2k + 2)=10(1.4k - 1)\), he should get \(2k+20 = 14k-10\) (using the distributive property \(a(b + c)=ab+ac\)). But he wrote \(2k + 2=14k - 1\), which means he didn't multiply 10 with 2 and - 1 correctly.

Step2: Solve the correct equation

Start with the correct equation \(0.2k+2 = 1.4k - 1\).
Subtract \(0.2k\) from both sides: \(2=1.4k-0.2k - 1\), so \(2 = 1.2k-1\).
Add 1 to both sides: \(2 + 1=1.2k\), then \(3 = 1.2k\).
Divide both sides by \(1.2\): \(k=\frac{3}{1.2}=\frac{30}{12}=\frac{5}{2}=2.5\).

Answer:

a. Vic did not use the distributive property correctly when multiplying \(10(0.2k + 2)\) and \(10(1.4k - 1)\). He should have \(2k + 20=14k-10\) instead of \(2k + 2=14k - 1\).
b. \(k = 2.5\)