What is the common solution for the equations y = 3x + 1 and y = x + 3? Write your answer an ordered pair!

(1, 4 )

Step-by-step explanation:

Given the 2 equations

y = 3x + 1 → (1)

y = x + 3 → (2)

Substitute y = 3x + 1 into (2)

3x + 1 = x + 3 ( subtract x from both sides )

2x + 1 = 3 ( subtract 1 from both sides )

2x = 2 ( divide both sides by 2 )

x = 1

Substitute x = 1 into either of the 2 equations and evaluate for y

Substituting into (2)

y = 1 + 3 = 4

solution is (1, 4 )

(1,4)

Step-by-step explanation:

Hi there!

We're given the system:

y=3x+1

y=x+3

We need to find the common solution (the point at which the lines intersect); we do this by finding the value of the variables.

Let's start with x

Since both equations equal y, we can make 3x+1 and x+3 equal to each other, because of a property known as transitivity (if a=b and b=c, then a=c)

3x+1=x+3

subtract 1 from both sides

3x=x+2

subtract x from both sides

2x=2

divide both sides by 2

x=1

we found the value of x

now we need to find the value of y

We do this by plugging x as 1 into either one of the equations to solve for y

let's use y=x+3 for example

substitute x as 1

y=1+3

add

y=4

we found the value of y.

Your question asks for the answer to be written as an ordered pair, which will be (1,4)

Hope this helps!

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !