Question

Rahul solved the equation 2(x – ) – 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction x = 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction . In which step did he use the addition property of equality? A table titled Rahul's Solution with 2 columns and 5 rows. The first column, Steps, has the entries 1, 2, 3, 4. The second column, Resulting equations, has the entries, 2 x minus StartFraction 1 Over 4 EndFraction minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x minus StartFraction 1 Over 4 EndFraction equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x equals StartFraction 56 Over 4 EndFraction, x equals 10. Step 1 Step 2 Step 3 Step 4

Answer 1

Answer:

c. step 3

Step-by-step explanation:

Answer 2

Answer:

Step 3

Step-by-step explanation:

The solution is given in the image attached. The steps are:

Step 1:

$2x-\frac{1}{4} -\frac{3}{5}x=\frac{55}{4}$

Step 2: simplifying the coefficients of x:

$2x -\frac{3}{5}x-\frac{1}{4}=\frac{55}{4}\\\frac{10x-3}{5} -\frac{1}{4}=\frac{55}{4}\\\frac{7x}{5} -\frac{1}{4}=\frac{55}{4}$

Step 3: Adding 1/4 to both sides

$\frac{7x}{5} -\frac{1}{4}+\frac{1}{4} =\frac{55}{4}+\frac{1}{4}\\ \frac{7x}{5}=\frac{55+1}{4}\\ \frac{7x}{5}=\frac{56}{4}$

Step 4: Multiplying both sides by 5/7

$\frac{7x}{5}*\frac{5}{7} =\frac{56}{4}*\frac{5}{7} \\x=10$

The addition property of equality states that if a number is added to both sides of an equation, the equation is still valid (i.e the equation is still the same). From the steps above, The addition property of equality was applied in step 3