Here are the solutions to the systems of equations using the substitution method.
1.
Given the system:
x+21y=5(1)
y=4x(2)
Step 1: Substitute equation (2) into equation (1).
x+21(4x)=5
x+2x=5
3x=5
x=35
Step 2: Substitute x=35 into equation (2).
y=4(35)
y=320
The solution is (x,y)=(35,320).
x=35,y=320
2.
Given the system:
x=21y+2(1)
x+y=10(2)
Step 1: Substitute equation (1) into equation (2).
(21y+2)+y=10
23y+2=10
23y=8
y=8×32
y=316
Step 2: Substitute y=316 into equation (1).
x=21(316)+2
x=38+36
x=314
The solution is (x,y)=(314,316).
x=314,y=316
3.
Given the system:
y=21x−1(1)
x+y=7(2)
Step 1: Substitute equation (1) into equation (2).
x+(21x−1)=7
23x−1=7
23x=8
x=8×32
x=316
Step 2: Substitute x=316 into equation (1).
y=21(316)−1
y=38−33
y=35
The solution is (x,y)=(316,35).
x=316,y=35
4.
Given the system:
x=52y(1)
x+y=9(2)
Step 1: Substitute equation (1) into equation (2).
52y+y=9
57y=9
y=9×75
y=745
Step 2: Substitute y=745 into equation (1).
x=52(745)
x=718
The solution is (x,y)=(718,745).
x=718,y=745
5.
Given the system:
y=43x(1)
x+y=14(2)
Step 1: Substitute equation (1) into equation (2).
x+43x=14
47x=14
x=14×74
x=8
Step 2: Substitute x=8 into equation (1).
y=43(8)
y=6
The solution is (x,y)=(8,6).
x=8,y=6
6.
Given the system:
x=21y+3(1)
x−y=1(2)
Step 1: Substitute equation (1) into equation (2).
(21y+3)−y=1
−21y+3=1
−21y=−2
y=4
Step 2: Substitute y=4 into equation (1).
x=21(4)+3
x=2+3
x=5
The solution is (x,y)=(5,4).
x=5,y=4
7.
Given the system:
y=32x+1(1)
x+y=11(2)
Step 1: Substitute equation (1) into equation (2).
x+(32x+1)=11
35x+1=11
35x=10