Question 104455
Use the substitution method to solve each system and provide the coordinates. I've got most of them, but these ones I keep getting wrong.  Can someone help
please?
<pre><font size = 4><b>
(1) {{{3x-2y=33/2}}}
    {{{x/2-11/4=2y}}}

First clear them both of fractions.  To clear the first equation of
fractions multiply every term by the LCD of 2

{{{2(3x)-2(2y) = 2(33/2)}}}

which simplifies to

{{{6x - 4y = 33}}} which has no fractions.

To clear the second equation of
fractions multiply every term by LCD of 4

{{{4(x/2)-4(11/4)=4(2y)}}}

which simplifies to

{{{2x - 11 = 8y}}} which has no fractions

So now the system of equations to solve is:

{{{6x - 4y = 33}}}
{{{2x - 11 = 8y}}}

Rearranging the second equation like the
first equation

{{{6x - 4y = 33}}}
{{{2x - 8y = 11}}}

To make the y's cancel out, multiply the
first equation through by -2

            {{{6x - 4y = 33}}}
{{{(-2)(6x) - (-2)(4y) = (-2)(33)}}}
which simplifies to
          {{{-12x + 8y = -66}}}

Now place the second equation directly under
this:

{{{-12x + 8y = -66}}}
 {{{ 2x - 8y = 11}}}

Adding term by term gives

{{{-12x + 8y = -66}}}
 {{{ 2x - 8y = 11}}}
--------------------
     {{{-10x = -55}}}
        {{{x = (-55)/(-10)}}}
        {{{x = 55/10}}}
        {{{x = 11/2}}}

To make the x's cancel out, multiply the
second equation through by -3.

{{{6x - 4y = 33}}}
{{{2x - 8y = 11}}}

{{{(-3)(2x) - (-3)(8y) = (-3)(11)}}}

{{{-6x + 24y = -33}}}

Now place this equation directly under the
first equation:

 {{{6x -  4y =  33}}}
{{{-6x + 24y = -33}}}

Adding term by term gives

 {{{6x -  4y =  33}}}
{{{-6x + 24y = -33}}}
--------------------
      {{{20y = 0}}}
        {{{y = 0/20}}}
        {{{y = 0}}}
       
So the solution is (x,y) = ({{{11/2}}},0)      

============================================= 

(2) {{{4x+5y+1=-10+2x}}}
    {{{x-3y+2=-1-x}}}  

Rearrange the terms of the first equation as {{{2x+5y=-11}}}
Rearrange the terms of the second equation as {{{-2x-3y=-1}}}

 {{{2x+5y=-11}}}
{{{-2x-3y=-1}}}

The x's will cancel as they are:

Add term by term:

 {{{2x+5y=-11}}}
{{{-2x-3y=-1}}}
-----------------
       2y=-12
     {{{y=(-12)/2}}} 
     {{{y=-6}}}

Substitute -6 for y in the 1st equation:

    {{{2x+5y=-11}}}
 {{{2x+5(-6)=-11}}}
    {{{2x-30=-11}}}
       {{{2x = 19}}} 
        {{{x = 19/2}}}

So the solution is (x,y) = ({{{19/2}}},-6)

=======================================

(3)  {{{r+3s=15}}}
    {{{3r+2s=17}}} 

To make the r's cancel, multiply the
first equation by -3

            {{{r+3s=15}}}
{{{(-3)(r)+(-3)(3s)=(-3)(15)}}}
          {{{-3r-9s=-45}}}
          
Now write the second equation under this:
 
          {{{-3r-9s=-45}}}
           {{{3r+2s=17}}} 
        
Add term by term:

          {{{-3r-9s=-45}}}
           {{{3r+2s=17}}}
         ------------------
             {{{-7s=-28}}}
               {{{s=(-28)/(-7)}}}  
               {{{s=4}}}

Substitute 4 for s in 
            {{{r+3s=15}}}     
          {{{r+3(4)=15}}}
            {{{r+12=15}}}
               {{{r=3}}}

So the solution is (r,s) = (3,4). So
yes it it is (3,4)!

Edwin</pre>