Question 1135198
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There are always dozens of different paths you can take to solve a system of equations like this.  One path I would not take is the one the other tutor showed, which leads you through a whole bunch of ugly fractions.<br>
In general, in my opinion, the easiest way to solve a system like this is to eliminate one variable at a time.  Eliminate one variable to give a system of two equations in two unknowns; then eliminate one of the remaining variables and solve for the other.  Then work back through your equations to solve for the other two variables.<br>
Even with that general method, there are multiple paths you can take.  Here is one....<br>
{{{5x - 6y + 4z = 15}}}
{{{7x + 4y - 3z = 19}}}
{{{2x + y + 6z = 46}}}<br>
The coefficients make it look as if eliminating either y or z first will be easiest; I chose to eliminate z.<br>
If I multiply the first equation by 3 and the second by 4, the coefficients of z will be 12 and -12; when I add the two resulting equations, z will be eliminated.<br>
Similarly if I multiply the second equation by 2 and add it to the third equation.<br>
{{{15x-18y+12z = 45}}}
{{{28x+16y-12z = 76}}}
{{{43x-2y = 121}}}  [1]<br>
{{{14x+8y-6z = 38}}}
{{{2x+y+6z = 46}}}
{{{16x+9y = 84}}}  [2]<br>
Now eliminate one of the variables between equations [1] and [2].  Eliminating y looks easier; multiply the first equation by 9 and the second by 2 and add.<br>
{{{387x-18y = 1089}}}
{{{32x+18y = 168}}}
{{{419x = 1257}}}
{{{x = 3}}}  [3]<br>
Now work backwards to solve for the other variables.<br>
Substitute [3] in [2]:<br>
{{{16*3+9y = 84}}}
{{{48+9y = 84}}}
{{{9y = 36}}}
{{{y = 4}}}  [4]<br>
And now substitute [3] and [4] in one of the original equations:<br>
{{{2(3)+4+6z = 46}}}
{{{10+6z = 46}}}
{{{6z = 36}}}
{{{z=6}}}  [5]<br>
The solution is x=3, y=4, z=6.