26. SOLVING A SYSTEM OF
LINEAR EQUATIONS
1) Must have both equations in the form ax + by = c :
"2x = !3y + 15 "2x + 3y = 15
%#
#
5x
=
2y
!
29
$
$5x ! 2y = !29
2) We want to eliminate one of the variables. To do this, we need to take
the first set of coefficients, flip them, and change the sign of one
number. Then, we multiply both equations by those numbers.
" 2x + 3y = 15
#
$ 5x ! 2y = !29
2
5
" 5(2x + 3y = 15)
#
$!2(5x ! 2y = !29)
5
2
flip them
change sign of
one number
5
–2
" 10x + 15y = 75
#
$!10x + 4 y = 58
19y = 133
19
19
y=7
3) Plug in the value of the variable you have just solved for into either of
the original equations and solve for the other variable. To check, plug
solutions for both variables into the remaining equation.
y = 7 ! 2x = "3(7) + 15
2x = "21+ 15
2x = "6
2
2
x = "3
Visit SHAHOMEWORK.com
Check
(!3,7)
5x = 2y ! 29
5(!3) = 2(7) ! 29
!15 = 14 ! 29
√
!15 = !15
Created by J. Shahom (June 2004)
Examples
1.
! a + 3b = 19 !1a + 3b = 19
$"
"
# a = 2b + 4
#1a % 2b = 4
!1(1a ! 2b = 4)
" 1a + 3b = 19
#
$!1a + 2b = !4
5b = 15
5 5
b=3
a + 3b = 19
a + 3(3) = 19
a + 9 = 19
!9 ! 9
a = 10
1(1a + 3b = 19)
(10,3)
Check
2.
a = 2b + 4
10 = 2(3) + 4
10 = 6 + 4
√
10 = 10
"2x = 14 ! 3y
"2x + 3y = 14
%#
#
$5x ! 2y = !3 $5x ! 2y = !3
" 10x + 15y = 70
#
$!10x + 4 y = 6
19y = 76
19 19
y=4
5(2x + 3y = 14)
!2(5x ! 2y = !3)
2x = 14 ! 3y
2x = 14 ! 3(4)
2x = 14 !12
2x = 2
2 2
x =1
(1,4)
Check
Visit SHAHOMEWORK.com
5x ! 2y = !3
5(1) ! 2(4) = !3
5 ! 8 = !3
√
!3 = !3
Created by J. Shahom (June 2004)