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Question 1 of 5
Solve the following simultaneous equations by substitution.
x-y=8x−y=8
2x+y=12x+y=1
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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.
x-yx−y |
== |
88 |
Equation 11 |
2x+y2x+y |
== |
11 |
Equation 22 |
Next, solve for xx in Equation 11.
x-yx−y |
== |
88 |
x-yx−y+y+y |
== |
88+y+y |
Add yy to both sides |
xx |
== |
8+y8+y |
Simplify |
Substitute xx into Equation 22.
22xx+y+y |
== |
11 |
Equation 22 |
22(8+y)(8+y)+y+y |
== |
11 |
x=8+yx=8+y |
16+2y+y16+2y+y |
== |
11 |
Distribute 22 inside the parenthesis |
16+3y16+3y |
== |
11 |
Simplify |
16+3y16+3y-16−16 |
== |
11-16−16 |
Solve for yy |
3y3y |
== |
-15−15 |
yy |
== |
-5−5 |
Divide both sides by 33 |
Now, substitute the value of yy into Equation 11
x-x−yy |
== |
88 |
Equation 11 |
x-x−(-5)(−5) |
== |
88 |
y=-5y=−5 |
x+5x+5 |
== |
88 |
x+5x+5-5−5 |
== |
88-5−5 |
Simplify |
xx |
== |
33 |
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Question 2 of 5
Solve the following simultaneous equations by substitution.
2a+b=52a+b=5
5a-3b=75a−3b=7
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The substitution method removes one of the variables by replacement in a pair of systems of equations.
First, label the two equations 11 and 22 respectively.
2a+b2a+b |
== |
55 |
Equation 11 |
5a-3b5a−3b |
== |
77 |
Equation 22 |
Next, solve for bb in Equation 11.
2a+b2a+b |
== |
55 |
2a+b2a+b-2a−2a |
== |
55-2a−2a |
Subtract 2a2a from both sides |
bb |
== |
5-2a5−2a |
Simplify |
Substitute bb into Equation 22.
5a-35a−3bb |
== |
77 |
Equation 22 |
5a-35a−3(5-2a)(5−2a) |
== |
77 |
b=5-2ab=5−2a |
5a-15+6a5a−15+6a |
= |
7 |
Distribute -3 inside the parenthesis |
11a-15 |
= |
7 |
Simplify |
11a-15+15 |
= |
7+15 |
Solve for a |
11a |
= |
22 |
a |
= |
2 |
Divide both sides by 11 |
Now, substitute the value of a into Equation 1
2a+b |
= |
5 |
Equation 1 |
2(2)+b |
= |
5 |
a=2 |
4+b |
= |
5 |
4+b-4 |
= |
5-4 |
Simplify |
b |
= |
1 |
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Question 3 of 5
Solve the following simultaneous equations by substitution.
x-y=1
2x+y=11
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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.
x-y |
= |
1 |
Equation 1 |
2x+y |
= |
11 |
Equation 2 |
Next, solve for x in Equation 1.
x-y |
= |
1 |
x-y +y |
= |
1 +y |
Add y to both sides |
x |
= |
1+y |
Simplify |
Substitute x into Equation 2.
2x+y |
= |
11 |
Equation 2 |
2(1+y)+y |
= |
11 |
x=1+y |
2y+2+y |
= |
11 |
Distribute 2 inside the parenthesis |
3y+2 |
= |
11 |
Simplify |
3y+2 -2 |
= |
11 -2 |
Solve for y |
3y |
= |
9 |
3y ÷3 |
= |
9 ÷3 |
Divide both sides by 3 |
y |
= |
3 |
Now, substitute the value of y into Equation 1
x-y |
= |
1 |
Equation 1 |
x-(3) |
= |
1 |
y=3 |
x-3 +3 |
= |
1 +3 |
Add 3 to both sides |
x |
= |
4 |
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Question 4 of 5
Solve the following simultaneous equations by substitution.
a-3b=2
2a-b=14
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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.
a-3b |
= |
2 |
Equation 1 |
2a-b |
= |
14 |
Equation 2 |
Next, solve for a in Equation 1.
a-3b |
= |
2 |
a-3b +3b |
= |
2 +3b |
Add 3b to both sides |
a |
= |
2+3b |
Simplify |
Substitute a into Equation 2.
2a -b |
= |
14 |
Equation 2 |
2(2+3b) -b |
= |
14 |
a=2+3b |
4+6b-b |
= |
14 |
Distribute 2 inside the parenthesis |
5b+4 |
= |
14 |
Simplify |
5b+4 -4 |
= |
14 -4 |
Solve for b |
5b |
= |
10 |
5b ÷5 |
= |
10 ÷5 |
Divide both sides by 5 |
b |
= |
2 |
Now, substitute the value of b into Equation 1
a-3b |
= |
2 |
Equation 1 |
a-3(2) |
= |
2 |
b=2 |
a-6 +6 |
= |
2 +6 |
Add 6 to both sides |
a |
= |
8 |
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Question 5 of 5
Solve the following simultaneous equations by substitution.
2x+5y=7
x-y=7
Incorrect
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Substitution Method
- 1) make one variable the subject
- 2) substitute into second equation
- 3) solve the second equation
- 4) substitute back
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.
2x+5y |
= |
7 |
Equation 1 |
x-y |
= |
7 |
Equation 2 |
Next, solve for x in Equation 2.
x-y |
= |
7 |
x-y +y |
= |
7 +y |
Add y from both sides |
x |
= |
7+y |
Simplify |
Substitute x into Equation 1.
2x +5y |
= |
7 |
Equation 1 |
2(7+y) +5y |
= |
7 |
x=7+y |
14+2y+5y |
= |
7 |
Distribute 2 inside the parenthesis |
14+7y |
= |
7 |
Simplify |
14+7y -14 |
= |
7 -14 |
Solve for y |
7y |
= |
-7 |
7y ÷7 |
= |
-7 ÷7 |
Divide both sides by 7 |
y |
= |
-1 |
Now, substitute the value of y into Equation 2
x- y |
= |
7 |
Equation 2 |
x- (-1) |
= |
7 |
y=-2 |
x+1 |
= |
7 |
Simplify |
x+1 -1 |
= |
7 -1 |
Subtract 1 from both sides |
x |
= |
6 |