SOLUTION: Solve using the substitution method.
#1
y=5-x
3x-4y=-20
#2
x+2y=6
2x+3y=8
#3
3x+y=1
x=2y+5
#4
x+y=6
y=3-2x
#5
s+t=5
s=13-3t
#6
x-y=4
y=2-x
Algebra.Com
Question 4990: Solve using the substitution method.
#1
y=5-x
3x-4y=-20
#2
x+2y=6
2x+3y=8
#3
3x+y=1
x=2y+5
#4
x+y=6
y=3-2x
#5
s+t=5
s=13-3t
#6
x-y=4
y=2-x
Answer by nidz(2) (Show Source): You can put this solution on YOUR website!
#1
y=5-x
3x-4y=-20
sol:
let,
y=5-x ..eq(i)
3x-4y=-20 ..eq(ii)
substitute the value of y from eq(i) in eq(ii),
then eq(ii) will be
3x-4(5-x)=-20
3x-20-4x=-20
-20-x=-20
x=-20+20
x=0
#2
x+2y=6
2x+3y=8
sol:
let,
x+2y=6 ..eq(i)
2x+3y=8 ..eq(ii)
from eq(i),
x=6-2y
substitute the value of x in eq(ii),
then eq(ii) will be
2(6-2y)+3y=8
12-4y+3y=8
12-y=8
y=12-8
y=4
#3
3x+y=1
x=2y+5
sol:
let,
3x+y=1 ..eq(i)
x=2y+5 ..eq(ii)
substitute the value of x from eq(ii) in eq(i),
then eq(i)will be
3(2y+5)+y=1
6y+15+y=1
7y+15=1
7y=1-15
7y=-14
y=-2
#4
x+y=6
y=3-2x
sol:
let,
x+y=6 ..eq(i)
y=3-2x ..eq(ii)
substitute the value of y from eq(ii) in eq(i)
then eq(i) will be
x+(3-2x)=6
x+3-2x=6
3-x=6
x=3-6
x=-3
#5
s+t=5
s=13-3t
sol:
let,
s+t=5 ..eq(i)
s=13-3t .eq(ii)
substitute the value of s from eq(ii) in eq(i)
(13-3t)+t=5
13-2t=5
13-5=2t
8=2t
t=4
#6
x-y=4
y=2-x
sol:
let,
x-y=4 ..eq(i)
y=2-x ..eq(ii)
substitute y in eq(i)
x-(2-x)=4
x-2+x=4
2x-2=4
2x=4+2
2x=6
x=3
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