Math

QuestionSolve for pp in the equation: 15p4(66p)=6(p3)3-15 p - 4(6 - 6 p) = 6(p - 3) - 3. What is the solution set? A, B, or C?

Studdy Solution

STEP 1

Assumptions1. We are solving for the variable pp in the equation 15p4(66p)=6(p3)3-15p -4(6 -6p) =6(p -3) -3. . The equation is a linear equation, and we assume that it has a solution.
3. The solution set is either a single real number, all real numbers (in case the equation is an identity), or no solution (in case the equation is a contradiction).

STEP 2

First, we need to simplify the equation by distributing the numbers outside the parentheses on both sides of the equation.
On the left side, distribute 4-4 across (66p)(6 -6p), and on the right side, distribute 66 across (p)(p -).
15p46+46p=6p6-15p -4 \cdot6 +4 \cdot6p =6 \cdot p -6 \cdot -

STEP 3

implify the equation by performing the multiplication.
15p24+24p=6p183-15p -24 +24p =6p -18 -3

STEP 4

Combine like terms on both sides of the equation.
9p24=6p219p -24 =6p -21

STEP 5

To isolate pp, we need to get all terms with pp on one side of the equation and the constants on the other side. Subtract pp from both sides to get the pp terms on the left side and add 2424 to both sides to get the constants on the right side.
9pp=21+249p -p =21 +24

STEP 6

implify the equation by performing the subtraction and addition.
3p=453p =45

STEP 7

Finally, to solve for pp, divide both sides of the equation by 33.
p=453p = \frac{45}{3}

STEP 8

implify the right side of the equation to find the value of pp.
p=15p =15

STEP 9

Now that we have the value of pp, we need to check our solution by substituting p=15p =15 back into the original equation.
15(15)4(66(15))=6(153)3-15(15) -4(6 -6(15)) =6(15 -3) -3

STEP 10

implify the equation by performing the multiplication.
2254(84)=723-225 -4(-84) =72 -3

STEP 11

Perform the multiplication and subtraction to check if both sides of the equation are equal.
225+336=69-225 +336 =69

STEP 12

implify both sides of the equation.
111=69111 =69

STEP 13

Since 11169111 \neq69, our solution p=15p =15 is incorrect. Therefore, the equation has no solution.
The solution set is \varnothing, so the equation is a contradiction.

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