Solve the following equation

x+5=12

Answers

Question 1

Carly and Janiya put some money into their money boxes every week. The amounts of money (y), in dollars, in their money boxes after certain amounts of time (x), in weeks, are shown by the equations below:

Carly

y = 60x + 40

Janiya

y = 50x + 80

After how many weeks will Carly and Janiya have the same amount of money in their money boxes and what will that amount be?

5 weeks, $280

5 weeks, $340

4 weeks, $280

4 weeks, $340

Question 2

A system of equations is shown below:

y = 8x − 2

y = 9x − 7

What is the solution to the system of equations?

(−5, 38)

(−5, −38)

(5, 38)

(5, −38)

Question 3

Two equations are given below:

m + 3n = 10

m = n − 2

What is the solution to the set of equations in the form (m, n)?

(1, 3)

(2, 4)

(0, 2)

(4, 6)

Question 4

The work of a student to solve a set of equations is shown:

Equation 1: m = 8 + 2n

Equation 2: 3m = 4 + 4n

Step 1:

−3(m) =

−3(8 + 2n)

[Equation 1 is multiplied by −3.]

3m =

4 + 4n

[Equation 2]

Step 2:

−3m =

−24 − 6n

[Equation 1 in Step 1 is simplified.]

3m =

4 + 4n

[Equation 2]

Step 3:

−3m + 3m =

−24 − 6n + 4n

[Equations in Step 2 are added.]

Step 4:

0 =

−24 − 2n

Step 5:

n =

−12

In which step did the student first make an error?

Step 4

Step 3

Step 2

Step 1

Question 5

A student is trying to solve the system of two equations given below:

Equation P: y + z = 6

Equation Q: 8y + 7z = 1

Which of the following is a possible step used in eliminating the y-term?

(y + z = 6) ⋅ −8

(y + z = 6) ⋅ 7

(8y + 7z = 1) ⋅ 7

(8y + 7z = 1) ⋅ 8

Question 6

Line A is represented by the equation given below:

x + y = 4

What is most likely the equation for line B, so that the set of equations has infinitely many solutions?

4x + 4y = 4

4x + y = 4

2x + 2y = 8

x + y = 8

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by EvilEvie 10.05.2018

Answers

SamanthaBees · Beginner

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MsEHolt Brainly Teacher

#1) 4 weeks, $280

#2) (5, 38)

#3) (1, 3)

#4) Step 3

#5) (y+z=6)*-8

#6) 2x+2y=8

Explanation

#1) Setting them equal,

60x+40=50x+80

Subtract 50x from both sides:

60x+40-50x=50x+80-50x

10x+40=80

Subtract 40 from both sides:

10x+40-40=80-40

10x=40

Divide both sides by 10:

10x/10 = 40/10

x=4

Plugging this in to one of our equations,

60(4)+40=240+40=280

#2) Setting the equations equal to one another,

8x-2=9x-7

Subtract 8x from both sides:

8x-2-8x=9x-7-8x

-2=x-7

Add 7 to both sides:

-2+7=x-7+7

5=x

Plugging this in to the first equation,

8(5)-2=y

40-2=y

38=y

#3) Substituting our value from the second equation into the first one,

n-2+3n=10

Combining like terms,

4n-2=10

Add 2 to both sides:

4n-2+2=10+2

4n=12

Divide both sides by 4:

4n/4 = 12/4

n=3

Substitute this into the second equation:

m=3-2=1

#4) The mistake was made on Step 3; the 4 was left off when the equations were added.

#5) To eliminate y, we want the coefficients to be the same. To accomplish this, we will multiply the first equation by -8.

#6) In order to have infinitely many solutions, we want each coefficient as well as the constant to be a multiple of our equation. Multiplying the equation by 2, we get 2x+2y=8.

7.The solution to the set of equation in the form (-5,-3).

8.Multiply equation A by -4 used to eliminate  the z- term.

9.Step 2: -4y=-16+8z { equation A in step1 is simplified}.

10. p= d+2

p=d-1.

Step-by-step explanation:

7. Two equation are given below:

a-3b=4

a=b-2

II eqaution can be write as

a-b=-2

Subtracting equation II from  equation I then we get

-2b=6

By division property of equality

b=\frac{6}{-2}

By simplification we get

b=-3

Substitute the value of b in equation I then we get

a-3(-3)=4

a+9=4

a=4-9

a=-5

Hence, the solution of the set of equation is (-5,-3).

8. Equation A: x+z=6

Equation B: 2x+4z=1

Equation A is multiplied by -4 then we get

Equation A:-4x-4z=-24

Adding both equation A and B then we get

-2x=-23

Multiply equation A by -4 to eliminate the z-term.

9.Equation A: y=4-2z

Equation B:4y=2-4z

Step1 :-4(y)=-4(4-2z)

Equation A is multiplied by -4

4y=2-4z [equation B]

Step 2: -4y=-16+8z

4y=2-4z [equation B]

Equation A in step1 s simplified .

Step3: 0= -14+4z

Equations in step 2 are added.

Step 4: 4z=14

Step5: z=\frac{7}{2}

Hence, in step 2 student did make first an error.

10. Given

Variable p is more than variable d

We can write in algebraic expression

p=d+2

Variable p is also 1 less than variable d.

Then the algebraic expression

p=d-1

Hence, p=d+2

p=d-1

7.The solution to the set of equation in the form (-5,-3).

8.Multiply equation A by -4 used to eliminate  the z- term.

9.Step 2: -4y=-16+8z { equation A in step1 is simplified}.

10. p= d+2

p=d-1.

Step-by-step explanation:

7. Two equation are given below:

a-3b=4

a=b-2

II eqaution can be write as

a-b=-2

Subtracting equation II from  equation I then we get

-2b=6

By division property of equality

b=\frac{6}{-2}

By simplification we get

b=-3

Substitute the value of b in equation I then we get

a-3(-3)=4

a+9=4

a=4-9

a=-5

Hence, the solution of the set of equation is (-5,-3).

8. Equation A: x+z=6

Equation B: 2x+4z=1

Equation A is multiplied by -4 then we get

Equation A:-4x-4z=-24

Adding both equation A and B then we get

-2x=-23

Multiply equation A by -4 to eliminate the z-term.

9.Equation A: y=4-2z

Equation B:4y=2-4z

Step1 :-4(y)=-4(4-2z)

Equation A is multiplied by -4

4y=2-4z [equation B]

Step 2: -4y=-16+8z

4y=2-4z [equation B]

Equation A in step1 s simplified .

Step3: 0= -14+4z

Equations in step 2 are added.

Step 4: 4z=14

Step5: z=\frac{7}{2}

Hence, in step 2 student did make first an error.

10. Given

Variable p is more than variable d

We can write in algebraic expression

p=d+2

Variable p is also 1 less than variable d.

Then the algebraic expression

p=d-1

Hence, p=d+2

p=d-1

7. b - 2 - 3b = 4
   -2b = 6
    b = -3
    a = -5
solution is (-5,-3) <==
8. multiply A by -4...this eliminates the z's when added <==
9. first error...step 2he didn't distribute correctly <==
10. p = d + 2 : p = d - 1 <==
    
1) 4 weeks and $280
2) (5, 38)
3) (1, 3)
4) step 3
5) (y+2=6) *-8
6) 2x+2y=8
#1) 4 weeks, $280
#2) (5, 38)
#3) (1, 3)
#4) Step 3
#5) (y+z=6)*-8
#6) 2x+2y=8

Explanation
#1) Setting them equal,
60x+40=50x+80

Subtract 50x from both sides:
60x+40-50x=50x+80-50x
10x+40=80

Subtract 40 from both sides:
10x+40-40=80-40
10x=40

Divide both sides by 10:
10x/10 = 40/10
x=4

Plugging this in to one of our equations,
60(4)+40=240+40=280

#2) Setting the equations equal to one another, 
8x-2=9x-7

Subtract 8x from both sides:
8x-2-8x=9x-7-8x
-2=x-7

Add 7 to both sides:
-2+7=x-7+7
5=x

Plugging this in to the first equation,
8(5)-2=y
40-2=y
38=y

#3) Substituting our value from the second equation into the first one,
n-2+3n=10

Combining like terms,
4n-2=10

Add 2 to both sides:
4n-2+2=10+2
4n=12

Divide both sides by 4:
4n/4 = 12/4
n=3

Substitute this into the second equation:
m=3-2=1

#4) The mistake was made on Step 3; the 4 was left off when the equations were added.

#5) To eliminate y, we want the coefficients to be the same.  To accomplish this, we will multiply the first equation by -8.

#6) In order to have infinitely many solutions, we want each coefficient as well as the constant to be a multiple of our equation.  Multiplying the equation by 2, we get 2x+2y=8.
X+5=12
5-X+5=12-5
X=7
THATS ALL YOU NEED TO DO
X+5=12
5-X+5=12-5
X=7
THATS ALL YOU NEED TO DO
The first thing you do is subtract negative 5 on both side because we want to get x by itself.
Then we subtract 12 -5 which get us 7.
So, the solution is x=7.
And that how you solve the equation.
The first thing you do is subtract negative 5 on both side because we want to get x by itself.
Then we subtract 12 -5 which get us 7.
So, the solution is x=7.
And that how you solve the equation.


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