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:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.