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LuckyWell [14K]
2 years ago
13

solve the equations below. make sure you say how many solutions each equation has(MCC.8.EE.7a) A. 5p-9=7p-19 b. 5p-9= 5p-19 C. 5

p-9=5p-9
Mathematics
1 answer:
Sedaia [141]2 years ago
6 0
<h3>Answer:</h3>
  • A) p = 5, one solution
  • B) no solutions
  • C) infinite solutions
<h3>Step-by-step explanation:</h3>

A) Add 19-5p to each side of the equation:

... 10 = 2p

... 5 = p . . . . . divide by the coefficient of p

B) Subtract 5p from both sides of the equation:

... -9 = -19 . . . . . there is <em>no value of p</em> that will make this true. (No solution.)

C) Subtract 5p from both sides of the equation:

... -9 = -9 . . . . . this is true for <em>every value of p</em>. (Infinite solutions.)

You might be interested in
What is X in the equation -15x+15(-4)=-30
love history [14]

<em><u>Answer: x=-2 *The answer should be the negative sign.*</u></em>

Step-by-step explanation:

distributive property: a(b+c)=ab+ac

remove parenthesis

-15x-15*4=-30

15*4=60

-15x-60=-30

add 60 both sides of an equation.

-15x-60+60=-30+60

simplify.

-15x=30

divide by -15 both sides of an equation.

-15x/-15=30/-15

simplify.

30/15=2

x=-2

Hope this helps!

Thanks!

Have a great day!

7 0
3 years ago
-25+t=-44 solve. Use the addition principle to find solution. T=
Taya2010 [7]

To find T, you add 25 to both sides, isolating the varible. This gives you T=-19

Hope this helped!

4 0
3 years ago
Marks ratio is 3 tablespoons sugar to 4 tablespoons Milk.Shari is using 6 tablespoons of sugar to 8 tablespoons of milk. Eve is
Ganezh [65]
The awnser is Shari is using the same ratio as Mark's
8 0
3 years ago
Can you please give me the answer
IrinaK [193]

Step-by-step explanation:

This will help okay! I promise you, well I hope .

6 0
3 years ago
PLEASE HELP , TRYING TO BE ON HONOR ROLL !
borishaifa [10]

Option A

The solution is (0, \frac{-1}{4})

<em><u>Solution:</u></em>

<em><u>Given system of equations are:</u></em>

3x + 6y = 1 ------ eqn 1

x - 4y = 1 ------ eqn 2

We have to find solution to system of equations

We can use substitution method

From eqn 2,

x = 1 + 4y -------- eqn 3

Substitute eqn 3 in eqn 1

3(1 + 4y) + 6y = 1

3 + 12y + 6y = 1

18y = 1 - 3

18y = -2

Divide both sides by 18

y = \frac{-2}{18}\\\\y = \frac{-1}{4}

Substitute the above value of y in eqn 3

x = 1 + 4 \times \frac{-1}{4}\\\\x = 1-1\\\\x = 0

Thus solution is (0, \frac{-1}{4})

8 0
2 years ago
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