Information
You have already completed the quiz before. Hence you can not start it again.
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
-
Question 1 of 4
Solve the following systems of equations by substitution.
b=2a+2b=2a+2
2a+5b=102a+5b=10
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitution Method
- 1)1) make one variable the subject
- 2)2) substitute into second equation
- 3)3) solve the second equation
- 4)4) substitute back
First, label the two equations 11 and 22 respectively.
bb |
== |
2a+22a+2 |
Equation 11 |
2a+5b2a+5b |
== |
1010 |
Equation 22 |
Next, substitute bb from Equation 11 into Equation 22.
2a+52a+5bb |
== |
1010 |
Equation 22 |
2a+52a+5(2a+2)(2a+2) |
== |
1010 |
b=2a+2b=2a+2 |
2a+10a+102a+10a+10 |
== |
1010 |
Distribute 55 inside the parenthesis |
12a+1012a+10 |
== |
1010 |
Simplify |
12a+1012a+10 -10−10 |
== |
1010 -10−10 |
Solve for aa |
12a12a |
== |
00 |
12a12a ÷12÷12 |
== |
00 ÷12÷12 |
Divide both sides by 1212 |
aa |
== |
00 |
Now, substitute the value of aa into Equation 11
bb |
== |
22aa+2+2 |
Equation 11 |
bb |
== |
22(0)(0)+2+2 |
a=0a=0 |
bb |
== |
0+20+2 |
bb |
== |
22 |
-
Question 2 of 4
Solve the following systems of equations by substitution.
2m+5n=52m+5n=5
m-5n=10m−5n=10
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitution Method
- 1)1) make one variable the subject
- 2)2) substitute into second equation
- 3)3) solve the second equation
- 4)4) substitute back
First, label the two equations 11 and 22 respectively.
2m+5n2m+5n |
== |
55 |
Equation 11 |
m-5nm−5n |
== |
1010 |
Equation 22 |
Next, solve for mm in Equation 22.
m-5nm−5n |
== |
1010 |
m-5nm−5n +5n+5n |
== |
1010 +5n+5n |
Add 5n5n to both sides |
mm |
== |
10+5n10+5n |
Simplify |
Substitute mm into Equation 11.
22mm +5n+5n |
== |
55 |
Equation 11 |
22(10+5n)(10+5n) +5n+5n |
== |
55 |
m=10+5nm=10+5n |
20+10n+5n20+10n+5n |
== |
55 |
Distribute 22 inside the parenthesis |
20+15n20+15n |
== |
55 |
Simplify |
20+15n20+15n -20−20 |
== |
55 -20−20 |
Solve for nn |
15n15n |
== |
-15−15 |
15n15n ÷15÷15 |
== |
-15−15 ÷15÷15 |
Divide both sides by 1515 |
nn |
== |
-1−1 |
Now, substitute the value of nn into Equation 22
m-5m−5nn |
== |
1010 |
Equation 11 |
m-5m−5(-1)(−1) |
== |
1010 |
n=-1n=−1 |
m+5m+5 -5−5 |
== |
1010 -5−5 |
Subtract 55 from both sides |
mm |
== |
55 |
-
Question 3 of 4
Solve the following systems of equations by substitution.
4x-5y=144x−5y=14
2x+3y=-42x+3y=−4
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitution Method
- 1)1) make one variable the subject
- 2)2) substitute into second equation
- 3)3) solve the second equation
- 4)4) substitute back
First, label the two equations 11 and 22 respectively.
4x-5y4x−5y |
== |
1414 |
Equation 11 |
2x+3y2x+3y |
== |
-4−4 |
Equation 22 |
Next, solve for xx in Equation 22.
2x+3y2x+3y |
== |
-4−4 |
2x+3y2x+3y -3y−3y |
== |
-4−4-3y−3y |
Subtract 3y3y from both sides |
2x2x ÷2÷2 |
== |
(-4-3y)(−4−3y) ÷2÷2 |
Divide both sides by 22 |
|
xx |
== |
-4-3y2−4−3y2 |
Simplify |
Substitute xx into Equation 11.
44xx -5y−5y |
== |
1414 |
Equation 11 |
|
44(-4-3y2)(−4−3y2) -5y−5y |
== |
1414 |
x=-4-3y2x=−4−3y2 |
|
(4(−4−3y2)×2)−(5y×2)(4(−4−3y2)×2)−(5y×2) |
== |
1414×2×2 |
Multiply all values by 22 |
|
4(-4-3y)-10y4(−4−3y)−10y |
== |
2828 |
-16-12y-10y−16−12y−10y |
== |
2828 |
Distribute 44 inside the parenthesis |
-16-22y−16−22y |
== |
2828 |
Simplify |
-16-22y−16−22y +16+16 |
== |
2828 +16+16 |
Solve for yy |
-22y−22y |
== |
4444 |
-22y−22y ÷-22÷−22 |
== |
4444 ÷-22÷−22 |
Divide both sides by -22−22 |
yy |
== |
-2−2 |
Now, substitute the value of yy into Equation 22
2x+32x+3yy |
== |
-4−4 |
Equation 22 |
2x+32x+3(-2)(−2) |
== |
-4−4 |
y=-2y=−2 |
2x-62x−6 +6+6 |
== |
-4−4 +6+6 |
Add 66 to both sides |
2x2x ÷2÷2 |
== |
22 ÷2÷2 |
Divide both sides by 22 |
xx |
== |
11 |
-
Question 4 of 4
Solve the following systems of equations by substitution.
7x-2y=67x−2y=6
3x+4y=22
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitution Method
- 1) make one variable the subject
- 2) substitute into second equation
- 3) solve the second equation
- 4) substitute back
First, label the two equations 1 and 2 respectively.
7x-2y |
= |
6 |
Equation 1 |
3x+4y |
= |
22 |
Equation 2 |
Next, solve for y in Equation 1.
7x-2y |
= |
6 |
7x-2y +2y |
= |
6 +2y |
Add 2y to both sides |
7x -6 |
= |
6+2y -6 |
Subtract 2y from both sides |
(7x-6) ÷2 |
= |
2y ÷2 |
Divide both sides by 2 |
|
7x-62 |
= |
y |
Simplify |
|
y |
= |
7x-62 |
Simplify |
Substitute y into Equation 2.
3x+4y |
= |
22 |
Equation 2 |
|
3x+4(7x-62) |
= |
22 |
y=7x-62 |
|
(3x×2)+(4(7x−62)×2) |
= |
22×2 |
Multiply all values by 2 |
|
6x+4(7x-6) |
= |
44 |
6x+28x-24 |
= |
44 |
Distribute 4 inside the parenthesis |
34x-24 |
= |
44 |
Simplify |
34x-24 +24 |
= |
44 +24 |
Solve for x |
34x |
= |
68 |
34x ÷34 |
= |
68 ÷34 |
Divide both sides by 34 |
x |
= |
2 |
Now, substitute the value of x into Equation 1
7x -2y |
= |
6 |
Equation 1 |
7(2) -2y |
= |
6 |
x=2 |
14-2y -14 |
= |
6 -14 |
Subtract 14 from both sides |
-2y ÷-2 |
= |
-8 ÷-2 |
Divide both sides by -2 |
y |
= |
4 |