Y = 2x + 4
and
y = 2x - 8

Solve the systems by substitution

-x - y - z  = -8        (1a)

-4x + 4y +5z = 7       (2a)

2x + 2x =  4         (3a)

since (3a) a is an equation solely in x make x the subject of the equation

  2 ( x + x) = 4

 divide both sides by 2

      (x + x)  = 2

            2x  = 2

              x  =  1    (3b)

  substitute x in (3b) for x in (1a)

          -(1) - y - z  = -8  

    make y the subject of the equation

              -y - z  =  -8 + 1

                    -y =  -7 + z

                     y = 7 - z   (1b)

 substitute y in (1b) for y in (2a) as well as the value of x ( x= 1)

        -4(1)+ 4( 7 - z ) +5z = 7

       - 4 + 28 - 4z + 5z  = 7

              -24 + z  = 7

                       z  = 7 + 24

                       z  = 31

By using (1a) solve for y

          -(1) - y - (31)= -8          

          -1  - y - 31 = -8        

                        y = -1 - 31 + 8

                        y = -24

∴ the solutions of the system are  x = 1 ; y = -24 ; z = 31  

Hope this helps

Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations 2x-y=8;2x+y=4 so that you

Step-by-step explanation:

it would be:

c.parallelogram.

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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