<span>This question is from textbook </span><span>1. Solve by substitution or elimination method:
3x – 2y = 26
-7x + 3y = -49
Eliminate x:
3x – 2y = 26
-7x + 3y = -49
the coefficients of x are 3 and -7.
Ignoring signs temporarily they are 3 and 7.
The least common multiple of 3 and 7 is 21.
If we multiply the first equation through by 7,
there will be a 21x where the 3x is now. If we
multiply the second equation by 3, there will be
a -21x where the -7x is. And 21x and -21x will
cancel out when we add equals to equals. So
let's multiply the first equation through by 7
and the second equation through by 3:
7[ 3x – 2y = 26]
3[-7x + 3y = -49]
21x - 14y = 182
-21x + 9y = -147
Draw a line underneath and add equals to equals
vertically:
21x - 14y = 182
-21x + 9y = -147
-------------------
0x - 5y = 35
-5y = 35
Divide both sides by -5
y = -7
Eliminate y:
3x – 2y = 26
-7x + 3y = -49
the coefficients of y are -2 and 3.
Ignoring signs temporarily they are 2 and 3.
The least common multiple of 2 and 3 is 6.
If we multiply the first equation through by 3,
there will be a -6y where the -2y is now. If we
multiply the second equation by 2, there will be
a 6y where the 3y is. And -6y and 6y will
cancel out when we add equals to equals. So
let's multiply the first equation through by 3
and the second equation through by 2:
3[ 3x – 2y = 26]
2[-7x + 3y = -49]
9x - 6y = 78
-14x + 6y = -98
Draw a line underneath and add equals to equals
vertically:
9x - 6y = 78
-14x + 6y = -98
-------------------
-5x + 0y = -20
-5x = -20
Divide both sides by -5
x = 4
So the solution is
(x,y) = (4,-7)
====================================
2. Solve by substitution or elimination method:
4x – 5y = 14
-12x + 15y = -42
Eliminate x:
4x – 5y = 14
-12x + 15y = -42
The coefficients of x are 4 and -12
Ignoring signs temporarily they are 4 and 12.
The least common multiple of 4 and 12 is 12.
If we multiply the first equation through by 3,
there will be a 12x where the 4x is now. We don't
need to do anything to the second equation for
the 12x and the -12x will cancel out when we add
equals to equals. So let's multiply the first
equation through by 7 and leave the second equation
as it is:
3[ 4x – 5y = 14]
-12x + 15y = -42
12x - 15y = 42
-12x + 15y = -42
Draw a line underneath and add equals to equals
vertically:
12x - 15y = 42
-12x + 15y = -42
-------------------
0x + 0y = 0
As you can see, any number may be substituted
for x and y and that will always be true, since
you'll always get 0 on the left and that will
always give 0 and 0 will always equal to 0.
So there are infinitely many solutions. This
sort of system is called "dependent".
We need go no further. We just state that
the system is dependent and has INFINITELY
MANY solutions.
===============================================
3. Solve by substitution or elimination method:
-2x + 6y = 1
10x – 30y = -15
Eliminate x:
-2x + 6y = 1
10x – 30y = -15
The coefficients of x are -2 and -10
Ignoring signs temporarily they are 2 and 10.
The least common multiple of 2 and 10 is 10.
If we multiply the first equation through by 5,
there will be a -10x where the -2x is now. We don't
need to do anything to the second equation for
the 10x and the -10x will cancel out when we add
equals to equals. So let's multiply the first
equation through by 5 and leave the second equation
as it is:
5[ -2x + 6y = 1]
10x - 30y = -15
-10x + 30y = 5
10x - 30y = -15
Draw a line underneath and add equals to equals
vertically:
-10x + 30y = 5
10x - 30y = -15
-------------------
0x + 0y = -10 </span>
