Question 167587
Hi, Hope I can help
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Solve each system of equations:
x+2y=12
3y-4z=25
x+6y+z=20
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There is no fast way to solving these problems, but I will show you, what I believe is the easiest way, how to solve these problems
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First we have to solve for a variable, we will need to solve for "y" in each equation, since they all have a variable "y" ( It usually doesn't matter, but in this case it does.
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First equation, x+2y=12. (solve for "y")
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{{{ x+2y=12 }}}, We can solve for "y", first we will move the "x" to the right side
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{{{ x+2y =12 }}} = {{{ x-x+2y =12-x }}} = {{{ 2y  =12 - x }}}, we can rearrange the right side
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{{{ 2y  =12 - x }}} = {{{ 2y  = -x+12 }}}, to solve for "y" we will divide each side by "2"
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{{{ 2y  = -x+12 }}} = {{{ 2y/2  = (-x+12)/2 }}} = {{{ y  = (-x+12)/2 }}}
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{{{ (-x+12)/2 }}} is our first answer
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Second equation solve for "y", 3y-4z=25
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{{{ 3y-4z=25 }}}, we need to move (-4z) to the right side
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{{{ 3y-4z=25 }}} = {{{ 3y-4z + 4z =25 + 4z }}} = {{{ 3y = 25 + 4z }}}, we can rearrange the right side
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{{{ 3y = 25 + 4z }}} = {{{ 3y = 4z + 25 }}}, to solve for "y" we will need to divide each side by "3"
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{{{ 3y = 4z + 25 }}} = {{{ 3y/3 = (4z + 25)/3 }}} = {{{ y = (4z + 25)/3 }}}
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{{{ (4z + 25)/3 }}} is your second answer
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Now let us solve the last equation for "y", x+6y+z=20
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{{{ x+6y+z=20 }}}, first we will move the "z" to the right side
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{{{ x+6y+z=20 }}} = {{{ x+6y+z-z=20-z }}} = {{{ x+6y=20-z }}}, now let's move the "x" to the right side
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{{{ x+6y=20-z }}} = {{{ x-x+6y=20-z-x }}}, {{{ 6y=20-z-x }}}, let us rearrange the right side
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{{{ 6y=20-z-x }}} = {{{ 6y= -x - z + 20 }}}, to solve for "y", we will divide both sides by "6"
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{{{ 6y= -x - z + 20 }}} = {{{ 6y/6= (-x - z + 20)/6 }}} = {{{ y = (-x - z + 20)/6 }}}
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{{{ (-x - z + 20)/6 }}} is our third answer
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Now let us put all of our answers side by side
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First, {{{ (-x+12)/2 }}}
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Second, {{{(4z + 25)/3 }}}
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Third, {{{ (-x - z + 20)/6 }}}
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Since all three answers are equal to "y" all of them are equal to each other, first we have to have both "x" and "z" in all equations
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First, {{{ (-x+12)/2 }}}, this has no "z's" so this equation would be {{{ (-x+0z+12)/2 }}}
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Second, {{{(4z + 25)/3 }}}, this has no "x's" so this equation would be {{{ (0x + 4z + 25)/3 }}}
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Third, {{{ (-x - z + 20)/6 }}}, this already has both "x" and "z"
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Here are the new equations
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First, {{{ (-x+0z+12)/2 }}}
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Second, {{{(0x + 4z + 25)/3 }}}
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Third, {{{ (-x - z + 20)/6 }}}
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All of them equal each other, first we will let the first two equations equal each other
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{{{ (-x+0z+12)/2 = (0x + 4z + 25)/3 }}}, we will use cross multiplication
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{{{ (-x+0z+12)/2 = (0x + 4z + 25)/3 }}} = {{{ highlight(-x+0z+12)/2 = (0x + 4z + 25)/highlight(3) }}} = {{{ (-x+0z+12)/highlight(2) = highlight(0x + 4z + 25)/3 }}}
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{{{ (-x+0z+12)(3) = (2)(0x + 4z + 25) }}}, we will rearrange the left side
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{{{ (-x+0z+12)(3) = (2)(0x + 4z + 25) }}} = {{{ (3)(-x+0z+12) = (2)(0x + 4z + 25) }}}, now we can use the distribution method to solve even more,
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{{{ (3)(-x+0z+12) = (2)(0x + 4z + 25) }}} = {{{ highlight(3)(highlight(-x)+0z+12) = highlight(2)(highlight(0x) + 4z + 25) }}} = {{{ highlight(3)(-x+highlight(0z)+12) = highlight(2)(0x + highlight(4z) + 25) }}} = {{{ highlight(3)(-x+0z+highlight(12)) = highlight(2)(0x + 4z + highlight(25)) }}}
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Don't forget to use the positive, and negative signs
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{{{ -3x + 0z + 36 = 0x + 8z + 50 }}}, we will move the (-3x) and "0z" to the right side
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{{{ -3x + 0z + 36 = 0x + 8z + 50 }}} = {{{ -3x + 3x + 0z -0z + 36 = 0x + 3x + 8z -0z + 50 }}} = {{{  36 = 3x + 8z + 50 }}}, now we will move the "50" to the left side
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{{{  36 = 3x + 8z + 50 }}} = {{{  36 - 50 = 3x + 8z + 50 - 50 }}} = {{{  -14 = 3x + 8z }}}, rearranging we get
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{{{  -14 = 3x + 8z }}} = {{{  3x + 8z = -14 }}}
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{{{  3x + 8z = -14 }}} is our first answer
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Now we can use the last two equations ( we can use the middle equation twice
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Second, {{{(0x + 4z + 25)/3 }}}
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Third, {{{ (-x - z + 20)/6 }}}
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 {{{ (0x + 4z + 25)/3 = (-x - z + 20)/6 }}}, we can use cross multiplication
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{{{ (0x + 4z + 25)/3 = (-x - z + 20)/6 }}} = {{{ highlight(0x + 4z + 25)/3 = (-x - z + 20)/highlight(6) }}} = {{{ (0x + 4z + 25)/highlight(3) = highlight(-x - z + 20)/6 }}}
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{{{ (0x + 4z + 25)(6) = (3)(-x-z+20) }}}, rearranging the left side
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{{{ (0x + 4z + 25)(6) = (3)(-x-z+20) }}} = {{{ (6)(0x + 4z + 25) = (3)(-x-z+20) }}}, we will use the distrubution method,
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{{{ highlight(6)(highlight(0x) + 4z + 25) = highlight(3)(highlight(-x)-z+20) }}} = {{{ highlight(6)(0x + highlight(4z) + 25) = highlight(3)(-x-highlight(z)+20) }}} = {{{ highlight(6)(0x + 4z + highlight(25)) = highlight(3)(-x-z+highlight(20)) }}}
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Remember the signs, {{{ 0x + 24z + 150 = -3x-3z+60 }}}, we will move "-3x" to the left side
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{{{ 0x + 24z + 150 = -3x-3z+60 }}} = {{{ 0x+3x + 24z + 150 = -3x+3x-3z+60 }}} = {{{ 3x + 24z + 150 = -3z+60 }}}, we will now move "-3z to the left side
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{{{ 3x + 24z + 150 = -3z+60 }}} = {{{ 3x + 24z+3z + 150 = -3z+3z+60 }}} = {{{ 3x + 27z + 150 = 60 }}}, we will lastly move "150" to the right side
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{{{ 3x + 27z + 150 = 60 }}} = {{{ 3x + 27z + 150 - 150  = 60 - 150 }}} = {{{ 3x + 27z = -90 }}}
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{{{ 3x + 27z = -90 }}} is the second answer, once again let us put our two answers side by side
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First, {{{  3x + 8z = -14 }}}
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Second, {{{ 3x + 27z = -90 }}}, we can reduce this equation by dividing each side by "3", {{{ 3x + 27z = -90 }}} = {{{ (3x + 27z)/3 = -90/3 }}} = {{{ (3x/3) + (27z/3) = -30 }}} = {{{ x + 9z = -30 }}}
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Our new equations are
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First, {{{  3x + 8z = -14 }}}
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Second, {{{ x + 9z = -30 }}}
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Now we need to solve for a variable again, doesn't matter which one, we will solve for "x"
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First, {{{  3x + 8z = -14 }}}, we need to move "8z" to the right side,
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{{{  3x + 8z = -14 }}} = {{{  3x + 8z - 8z =  -14-8z }}} = {{{  3x =  -14-8z }}}
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Rearranging the right side, {{{  3x =  -14-8z }}} = {{{  3x =  -8z-14 }}}, we can solve for "x" by dividing each side by "3"
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{{{  3x =  -8z-14 }}} = {{{  3x/3 =  (-8z-14)/3 }}} = {{{  x =  (-8z-14)/3 }}}
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{{{  (-8z-14)/3 }}} is our first answer
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We can now solve for "x" in the second equation, {{{ x + 9z = -30 }}}
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{{{ x + 9z = -30 }}}, we will move "9z" to the right side
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{{{ x + 9z = -30 }}} = {{{ x + 9z-9z = -30-9z }}} = {{{ x = -30-9z }}}, rearranging the right side, {{{ x = -30-9z }}} = {{{ x = -9z-30}}}
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{{{ -9z-30}}} is our second answer, let us put our two answers together
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First, {{{  (-8z-14)/3 }}}
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Second, {{{ -9z-30}}}
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Since our two answers equal "x" they will equal each other
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 {{{  (-8z-14)/3 = -9z-30 }}} = {{{  (-8z-14)/3 = (-9z-30)/1 }}}, using cross-multiplication
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{{{  highlight(-8z-14)/3 = (-9z-30)/highlight(1) }}} = {{{  (-8z-14)/highlight(3) = highlight(-9z-30)/1 }}}
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{{{ (-8z-14)(1) = (3)(-9z-30) }}}, rearranging the left side
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{{{ (-8z-14)(1) = (3)(-9z-30) }}} = {{{ (1)(-8z-14) = (3)(-9z-30) }}}, using distribution
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{{{ highlight(1)(highlight(-8z)-14) = highlight(3)(highlight(-9z)-30) }}} = {{{ highlight(1)(-8z-highlight(14)) = highlight(3)(-9z-highlight(30)) }}}
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Remember the signs, {{{ -8z - 14 = -27z - 90 }}}, we will now solve for "z", we will move (-27z) to the left side
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{{{ -8z - 14 = -27z - 90 }}} = {{{ -8z+27z - 14 = -27z+ 27z - 90 }}} = {{{ 19z - 14 =  - 90 }}}, now we will move (-14) to the right side
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{{{ 19z - 14 =  - 90 }}} = {{{ 19z - 14+14 =  - 90+14 }}} = {{{ 19z =  - 76 }}}, to solve "z" we will divide each side by "19"
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{{{ 19z =  - 76 }}} = {{{ 19z/19 =  - 76/19 }}} = {{{ z =  - 4 }}}, to check our answer we will replace "z" with (-4) in our equation
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{{{  (-8z-14)/3 = -9z-30 }}}  = {{{  (-8(-4)-14)/3 = -9(-4)-30 }}} = {{{  ((32)-14)/3 = (36)-30 }}} = {{{  (18)/3 = (6) }}} = {{{  (6) = (6) }}} (True)
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Now that we know that {{{ z = -4 }}}, we can replace it in one of the equations with two variables
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First, {{{  3x + 8z = -14 }}}
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Second, {{{ x + 9z = -30 }}}
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We will use the second equation, replace "z" with (-4)
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{{{ x + 9z = -30 }}} = {{{ x + 9(-4) = -30 }}} = {{{ x + (-36) = -30 }}} = {{{ x  -36  = -30 }}}, lets move (-36) to the right side
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{{{ x  -36  = -30 }}} = {{{ x  -36+36  = -30+36 }}} = {{{ x  = 6 }}}
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Let's check our answer by replacing "x" with "6", and "z" with (-4) in the equation,
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{{{ x + 9z = -30 }}} = {{{ (6) + 9(-4) = -30 }}} = {{{ 6 + (-36) = -30 }}} = {{{ 6 - 36 = -30 }}} = {{{ -30 = -30 }}} (True)
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"x" = "6"
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"z" = (-4),
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Now we can find "y" by replacing "x" and "y" in one of the three original equations
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x+2y=12
3y-4z=25
x+6y+z=20
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We will need to use the third equation, replace "x" with "6", "z" with (-4)
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{{{ x+6y+z=20 }}} = {{{ (6)+6y+(-4)=20 }}} = {{{ (6)+6y- 4=20 }}} = {{{ 6-4+6y=20 }}} = {{{ 2+6y=20 }}}, we will move "2" to the right side
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{{{ 2+6y=20 }}} = {{{ 2-2+6y=20-2 }}} = {{{ 6y= 18 }}}, to find "y", divide each side by "6"
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{{{ 6y= 18 }}} = {{{ 6y/6= 18/6 }}} = {{{ y = 3 }}}, we can check by replacing "x" with "6", "y" with "3", "z" with (-4) in the equation
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{{{ x+6y+z=20 }}} = {{{ (6)+6(3)+(-4)=20 }}} = {{{ 6+18-4=20 }}} = {{{ 24-4=20 }}} = {{{ 20=20 }}} ( True )
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x = 6
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y = 3
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z = (-4)
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We can check by replacing the other two original equations
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{{{ x+2y=12 }}} = {{{ (6)+2(3)=12 }}} = {{{ 6+6=12 }}} = {{{ 12 = 12 }}} (True)
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{{{ 3y-4z=25 }}} = {{{ 3(3)-4(-4)=25 }}} = {{{ 9+ 16=25 }}} = {{{ 25 = 25 }}} (True)
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x = 6
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y = 3
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z = (-4)
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ordered pairs are given as (x,y,z), our ordered pair = ( 6,3,-4)
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Hope I helped, Levi