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 simultaneous equations by substitution.
8a+5b=68a+5b=6
4a+2b=44a+2b=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.
8a+5b8a+5b |
== |
66 |
Equation 11 |
4a+2b4a+2b |
== |
44 |
Equation 22 |
Next, solve for bb in Equation 22.
4a+2b4a+2b |
== |
44 |
(4a(4a÷2÷2)+(2b)+(2b÷2÷2)) |
== |
44÷2÷2 |
Divide the values of both sides by 22 |
2a+b2a+b -2a−2a |
== |
22 -2a−2a |
Subtract 2a2a from both sides |
bb |
== |
2-2a2−2a |
Simplify |
Substitute bb into Equation 11.
8a+58a+5bb |
== |
66 |
Equation 11 |
8a+58a+5(2-2a)(2−2a) |
== |
66 |
b=2-2ab=2−2a |
8a+10-10a8a+10−10a |
== |
66 |
Distribute 55 inside the parenthesis |
10-2a10−2a |
== |
66 |
Simplify |
10-2a10−2a -10−10 |
== |
66 -10−10 |
Solve for aa |
-2a−2a |
== |
-4−4 |
-2a−2a ÷(-2)÷(−2) |
== |
-4−4 ÷(-2)÷(−2) |
Divide both sides by -2−2 |
aa |
== |
22 |
Now, substitute the value of aa into Equation 22
44aa +2b+2b |
== |
44 |
Equation 11 |
44(2)(2) +2b+2b |
== |
44 |
a=2a=2 |
8+2b8+2b -8−8 |
== |
44 -8−8 |
Subtract 88 from both sides |
2b2b÷2÷2 |
== |
-4−4÷2÷2 |
Divide both sides by 22 |
bb |
== |
-2−2 |
-
Question 2 of 4
Solve the following simultaneous equations by substitution.
2x+5y=172x+5y=17
3x+3y=123x+3y=12
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.
2x+5y2x+5y |
== |
1717 |
Equation 11 |
3x+3y3x+3y |
== |
1212 |
Equation 22 |
Next, solve for xx in Equation 22.
3x+3y3x+3y |
== |
1212 |
(3x(3x÷3÷3)+(3y)+(3y÷3÷3)) |
== |
1212÷3÷3 |
Divide the values of both sides by 33 |
x+yx+y -y−y |
== |
44 -y−y |
Subtract yy from both sides |
xx |
== |
4-y4−y |
Simplify |
Substitute xx into Equation 11.
22xx +5y+5y |
== |
1717 |
Equation 11 |
22(4-y)(4−y) +5y+5y |
== |
1717 |
x=4-yx=4−y |
8-2y+5y8−2y+5y |
== |
1717 |
Distribute 22 inside the parenthesis |
8+3y8+3y |
== |
1717 |
Simplify |
8+3y8+3y -8−8 |
== |
1717 -8−8 |
Solve for yy |
3y3y |
== |
99 |
3y3y ÷3÷3 |
== |
99 ÷3÷3 |
Divide both sides by 33 |
yy |
== |
33 |
Now, substitute the value of yy into Equation 22
3x+33x+3yy |
== |
1212 |
Equation 22 |
3x+33x+3(3)(3) |
== |
1212 |
y=3y=3 |
3x+93x+9 -9−9 |
== |
1212 -9−9 |
Subtract 99 to both sides |
3x3x ÷3÷3 |
== |
33 ÷3÷3 |
Divide both sides by 33 |
xx |
== |
11 |
-
Question 3 of 4
Solve the following simultaneous equations by substitution.
3x+7y=43x+7y=4
2x+3y=12x+3y=1
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.
3x+7y3x+7y |
== |
44 |
Equation 11 |
2x+3y2x+3y |
== |
11 |
Equation 22 |
Next, solve for xx in Equation 22.
2x+3y2x+3y |
== |
11 |
2x+3y2x+3y -3y−3y |
== |
11-3y−3y |
Subtract 3y3y from both sides |
2x2x ÷2÷2 |
== |
(1-3y)(1−3y) ÷2÷2 |
Divide both sides by 22 |
|
xx |
== |
1-3y21−3y2 |
Simplify |
Substitute xx into Equation 11.
33xx +7y+7y |
== |
44 |
Equation 11 |
|
33(1-3y2)(1−3y2) +7y+7y |
== |
44 |
x=1-3y2x=1−3y2 |
|
(3(1−3y2)×2)+(7y×2)(3(1−3y2)×2)+(7y×2) |
== |
44×2×2 |
Multiply all values by 22 |
|
3(1-3y)+14y3(1−3y)+14y |
== |
88 |
3-9y+14y3−9y+14y |
== |
88 |
Distribute 33 inside the parenthesis |
3+5y3+5y |
== |
88 |
Simplify |
3+5y3+5y -3−3 |
== |
88 -3−3 |
Solve for yy |
5y5y |
== |
55 |
5y5y ÷5÷5 |
== |
55 ÷5÷5 |
Divide both sides by 55 |
yy |
== |
11 |
Now, substitute the value of yy into Equation 22
2x+32x+3yy |
== |
11 |
Equation 22 |
2x+32x+3(1)(1) |
== |
11 |
y=1y=1 |
2x+32x+3 -3−3 |
== |
11 -3−3 |
Subtract 33 from both sides |
2x2x ÷2÷2 |
== |
-2−2 ÷2÷2 |
Divide both sides by 22 |
xx |
== |
-1−1 |
-
Question 4 of 4
Solve the following simultaneous equations by substitution.
5a+2b=45a+2b=4
2a-3b=132a−3b=13
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.
5a+2b5a+2b |
== |
44 |
Equation 11 |
2a-3b2a−3b |
== |
1313 |
Equation 22 |
Next, solve for aa in Equation 22.
2a-3b2a−3b |
== |
1313 |
2a-3b2a−3b +3b+3b |
== |
1313+3b+3b |
Add 3b3b to both sides |
2a2a ÷2÷2 |
== |
(13+3b)(13+3b) ÷2÷2 |
Divide both sides by 22 |
|
aa |
== |
13+3b213+3b2 |
Simplify |
Substitute aa into Equation 11.
55aa +2b+2b |
== |
44 |
Equation 11 |
|
55(13+3b2)(13+3b2) +2b+2b |
== |
44 |
a=13+3b2a=13+3b2 |
|
(5(13+3b2)×2)+(2b×2)(5(13+3b2)×2)+(2b×2) |
== |
44×2×2 |
Multiply all values by 22 |
|
5(13+3b)+4b5(13+3b)+4b |
== |
88 |
65+15b+4b65+15b+4b |
== |
88 |
Distribute 55 inside the parenthesis |
65+19b65+19b |
== |
88 |
Simplify |
65+19b65+19b -65−65 |
== |
88 -65−65 |
Solve for bb |
19b19b |
== |
-57−57 |
19b19b ÷19÷19 |
== |
-57−57 ÷19÷19 |
Divide both sides by 1919 |
bb |
== |
-3−3 |
Now, substitute the value of bb into Equation 22
2a-32a−3bb |
== |
1313 |
Equation 2 |
2a-3(-3) |
= |
13 |
b=-3 |
2a+9 -9 |
= |
13 -9 |
Subtract 9 from both sides |
2a ÷2 |
= |
4 ÷2 |
Divide both sides by 2 |
a |
= |
2 |