Find four solutions for each equation. Write the solutions as ordered pair. y=x-5 and y=-3x+1

Your solution is (3/2, -7/2)

Step-by-step explanation:

y = x - 5

y = -3x + 1

First, see that these two equations equal y. This means you can set them equal to each other.

x - 5 = -3x + 1

Combine like terms.

[add 3x to both sides]

4x -5 = 1

[add 5 to both sides]

4x = 6

Isolate x further by dividing both sides by 4.

x = 6/4

Simplify the fraction.

x = 3/2

Now that we know x, plug its value (3/2) back into one of the original equations to find y.

y = (3/2) - 5

Simplify by subtracting.

[convert 5 to have a denominator of 2]

y = 3/2 - 10/2

y = -7/2

Your solution is (3/2, -7/2)

Check this by plugging these values into the equation you have not yet checked.

-7/2 = -3(3/2) + 1

Simplify.

[multiply]

-7/2 = -9/2 + 1

[convert 1 to have a denominator of 2]

-7/2 = -9/2 + 2/2

[add]

-7/2 = -7/2

This equation is true; therefore your solution is correct.

Hope this helps!

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2