Question 170002
Hi, Hope I can help,
.
Solve by substitution or elimination method:
.
3x - 2y = 8
-12x + 8y = 32
.
Solve by substitution or elimination method:
. 
7x - 5y = 14
-4x + y = 27
.
Solve by substitution or elimination method:
. 
-4x + 3y = 5
12x - 9y = -15
.
First we will solve the first system with substitution
.
3x - 2y = 8
-12x + 8y = 32
.
First we need to solve for a variable in one of the two equations, doesn't matter which letter, or equation, we will solve for "y" in the first equation
.
{{{ 3x - 2y = 8 }}}, we will move (-2y) to the right side
.
{{{ 3x - 2y = 8 }}} = {{{ 3x - 2y + 2y = 8 + 2y }}} = {{{ 3x = 8 + 2y }}}, now we need to move "8" to the left side
.
{{{ 3x = 8 + 2y }}} = {{{ 3x - 8 = 8 - 8 + 2y }}} = {{{ 3x - 8 = 2y }}}, to find "y" we need to divide each side by "2"
.
{{{ 3x - 8 = 2y }}} = {{{ (3x - 8)/2 = 2y/2 }}} = {{{ (3x - 8)/2 = y }}}
.
{{{ y = (3x - 8)/2 }}}, since "y" is equal to {{{ (3x - 8)/2 }}}, we can replace "y" in the other equation with {{{ (3x - 8)/2 }}}, then just solve for "x"
.
{{{ -12x + 8y = 32 }}} = {{{ -12x + 8((3x - 8)/2) = 32 }}}, now just solve for "x"
.
{{{ -12x + 8((3x - 8)/2) = 32 }}} = {{{ -12x + 4((3x - 8)/1) = 32 }}} = {{{ -12x + 4(3x - 8) = 32 }}}, now we will use the distribution method
.
{{{ -12x + highlight(4)(highlight(3x) - 8) = 32 }}} = {{{ -12x + highlight(4)(3x - highlight(8)) = 32 }}}
.
Remember the + and - signs,  {{{ -12x + 12x - 32 = 32 }}}, adding the "x"'s
.
{{{ -12x + 12x - 32 = 32 }}} = {{{ 0x - 32 = 32 }}} = {{{ - 32 = 32 }}} ( False )
.
This means ( when the x's or y's cancel out, and there is a false statement) that there are no solutions, these lines are parallel, there is no intersection, and therefore no solutions
.
Here is the graph of this system
.
{{{ graph ( 300,300,-10,10,-10,10, (3x - 8)/2, (3x+8)/2 ) }}}
.
Now we will solve the second system, by elimination
.
7x - 5y = 14
-4x + y = 27
.
Elimination is when you add the two equations together, and it gets rid of a variable, first we need to make sure the x's or y's in each equation are the same, or the negative of the other
.
We can eliminate any variable ( either x or y )
.
We will get rid of the y's
.
We need to either get both y's to "5y" or "-5y" or we need to get them to "y" or "-y", we will change the second equation to "5y"
.
{{{ -4x + y = 27 }}}, to get "y" to change to "5y" we need to multiply each side by (5)
.
{{{ -4x + y = 27 }}} = {{{ 5(-4x + y) = 5(27) }}} = {{{ 5(-4x + y) = 135 }}}
.
We will need to use the distribution method
.
{{{ 5(-4x + y) = 135 }}} = {{{ highlight(5)(highlight(-4x) + y) = 135 }}} = {{{ highlight(5)(-4x + highlight(y)) = 135 }}}
.
Remember the signs, {{{ -20x + 5y = 135 }}}, this is our new equation
.
Now we will bring the firt equation to our second new equation
.
{{{ 7x - 5y = 14 }}}
.
{{{ -20x + 5y = 135 }}}
.
We will now add the equations
.
{{{ 7x - 5y = 14 }}}
.
{{{ -20x + 5y = 135 }}}
.
{{{ 7x + (-20x) }}} = {{{ -13x }}}
.
{{{ -5y + 5y }}} = {{{ 0y }}} = {{{ 0 }}}
.
{{{ 14 + 135 }}} = {{{149}}}
.
It will become {{{ -13x + 0 = 149 }}} = {{{ -13x = 149 }}}, to find "x" we need to divide each side by {{{ -13 }}}
.
{{{ -13x = 149 }}} = {{{ -13x/-13 = 149/-13 }}} = {{{ x = 149/-13 }}} = {{{ x = -149/13 }}}, we can now replace "x" with {{{  -149/13 }}}, in one of the two original equations
.
7x - 5y = 14
-4x + y = 27
.
We will use the second equation
.
{{{ -4x + y = 27 }}} = {{{ -4(-149/13) + y = 27 }}} = {{{ (-4/1)(-149/13) + y = 27 }}} = {{{ (596/13) + y = 27 }}}
.
Now we need to move {{{ 596/13 }}} to the right side ( we will convert "27" into "13ths"
.
{{{ 596/13 + y = 27 }}} = {{{ 596/13 + y = 351/13 }}} = {{{ (596/13) - (596/13) + y = (351/13)-(596/13) }}} = {{{ y = (351/13)-(596/13) }}} = {{{ y = -245/13 }}}
.
{{{ y = -245/13 }}}, we can check our answers by replacing "x" and "y" in both  original equations
.
{{{ x = -149/13 }}}
.
{{{ y = -245/13 }}}
.
First equation, {{{ 7x - 5y = 14 }}} = {{{ 7(-149/13) - 5(-245/13) = 14 }}} = {{{ (7/1)(-149/13) - ((5/1)(-245/13)) = 14 }}} = {{{ (-1043/13) - (-1225/13) = 14 }}} = {{{ (-1043/13) + 1225/13 = 14 }}} = {{{ 182/13 = 14 }}} = {{{ 14 = 14 }}} ( True )
.
Second equation, {{{ -4x + y = 27 }}} = {{{ -4(-149/13) + (-245/13) = 27 }}} = {{{ (-4/1)(-149/13) - (245/13) = 27 }}} = {{{ (596/13) - (245/13) = 27 }}} = {{{ 351/13 = 27 }}} = {{{ 27 = 27 }}}, ( True )
.
{{{ x = -149/13 }}}
.
{{{ y = -245/13 }}}
.
Solution sets are in the form (x,y), our solution set is ( {{{ -149/13 }}}, {{{ -245/13 }}} )
.
The graph of the system is
.
{{{ drawing ( 500,500,-20,20,-20,20,red (circle (-149/13,-245/13,0.1)),blue (circle (-149/13,-245/13,0.2)),graph ( 500,500,-20,20,-20,20, (7x-14)/5, 4x+27)) }}}
.
The intersection is your answer
.
Now we can do the last system
.
-4x + 3y = 5
12x - 9y = -15
.
We will use the elimination method of solving this problem
.
Remember elimination is where you get rid of a variable
.
We will multiply the first equation by "3" to get rid of the "y"
.
{{{ -4x + 3y = 5 }}} = {{{ 3(-4x + 3y) = 3(5) }}} = {{{ 3(-4x + 3y) = 15 }}} we will use distribution
.
{{{ 3(-4x + 3y) = 15 }}} = {{{ highlight (3)(highlight(-4x) + 3y) = 15 }}} = {{{ highlight (3)(-4x + highlight(3y)) = 15 }}}
.
Remember signs, {{{ -12x + 9y = 15 }}}, now we can add the new first equation to the second equation
.
{{{ -12x + 9y = 15 }}}
.
{{{ 12x - 9y = -15 }}}
.
{{{ -12x + 12x }}} = {{{ 0x }}} = {{{ 0 }}}
.
{{{ 9y + (-9y) }}} = {{{ 9y - 9y }}} = {{{ 0y }}} = {{{ 0 }}}
.
{{{ 15 + (-15) }}} = {{{ 15 - 15 }}} = {{{ 0 }}}
.
It will become {{{ 0 + 0 = 0 }}} = {{{ 0 = 0 }}} (True)
.
This means ( when both x and y cancel out, and there is a true statement) that this is the same line, both equation will be one line, there is an infinite number of solutions ( since the equations are equal to one line )( or the second line is put on the first line )
.
Here is the graph of this system
.
{{{ graph ( 300,300, -10,10,-10,10, (4x + 5)/3, (4x+5)/3 ) }}}
.
Your answers are
.
First system of equations = " no solutions " ( since they are parallel lines )
.
Second system of equations = ( {{{ -149/13 }}}, {{{ -245/13 }}} ) ( the lines intersect at point ( {{{ -149/13 }}}, {{{ -245/13 }}} )
.
Third system of equations = " infinite solutions " ( since the two equations are the same line )
.
Hope I helped, Levi