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Question 1 of 5
Solve the following simultaneous equations by substitution.
2x-y=62x−y=6
x+3y=10x+3y=10
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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.
2x-y2x−y |
== |
66 |
Equation 11 |
x+3yx+3y |
== |
1010 |
Equation 22 |
Next, solve for yy in Equation 11.
2x-y2x−y |
== |
66 |
2x-y2x−y+y-6+y−6 |
== |
66+y-6+y−6 |
Add y-6y−6 to both sides |
2x-62x−6 |
== |
yy |
Simplify |
yy |
== |
2x-62x−6 |
Simplify |
Substitute yy into Equation 22.
x+3x+3yy |
== |
1010 |
Equation 22 |
x+3x+3(2x-6)(2x−6) |
== |
1010 |
y=2x-6y=2x−6 |
x+6x-18x+6x−18 |
== |
1010 |
Distribute 33 inside the parenthesis |
7x-187x−18 |
== |
1010 |
Simplify |
7x-187x−18+18+18 |
== |
1010+18+18 |
Solve for xx |
7x7x |
= |
28 |
x |
= |
4 |
Divide both sides by 7 |
Now, substitute the value of x into Equation 1
2x-y |
= |
6 |
Equation 1 |
2(4)-y |
= |
6 |
x=4 |
8-y |
= |
6 |
8-y-8 |
= |
6-8 |
Simplify |
-y |
= |
-2 |
y |
= |
2 |
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Question 2 of 5
Solve the following simultaneous equations by substitution.
2x+y=1
x-y=2
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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 1 and 2 respectively.
2x+y |
= |
1 |
Equation 1 |
x-y |
= |
2 |
Equation 2 |
Next, solve for x in Equation 2.
x-y |
= |
2 |
x-y+y |
= |
2+y |
Add y to both sides of the equation |
x |
= |
2+y |
Simplify |
Substitute x into Equation 1.
2x+y |
= |
1 |
Equation 1 |
2(2+y)+y |
= |
1 |
x=2+y |
4+2y+y |
= |
1 |
Distribute 2 inside the parenthesis |
4+3y |
= |
1 |
Simplify |
4+3y-4 |
= |
1-4 |
Solve for y |
3y |
= |
-3 |
y |
= |
-1 |
Divide both sides by 3 |
Now, substitute the value of y into Equation 2
x-y |
= |
2 |
Equation 2 |
x-(-1) |
= |
2 |
y=-1 |
x+1 |
= |
2 |
x+1-1 |
= |
2-1 |
Simplify |
x |
= |
1 |
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Question 3 of 5
Solve the following simultaneous equations by substitution.
2x+y=7
4x-2y=10
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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.
2x+y |
= |
7 |
Equation 1 |
4x-2y |
= |
10 |
Equation 2 |
Next, solve for y in Equation 1.
2x+y |
= |
7 |
2x+y -2x |
= |
7 -2x |
Subtract 2x from both sides |
y |
= |
7-2x |
Simplify |
Substitute y into Equation 2.
4x-2 y |
= |
10 |
Equation 2 |
4x-2 (7-2x) |
= |
11 |
y=7-2x |
4x-14+4x |
= |
10 |
Distribute 2 inside the parenthesis |
8x-14 |
= |
10 |
Simplify |
8x-14 +14 |
= |
10 +14 |
Solve for x |
8x |
= |
24 |
8x ÷8 |
= |
24 ÷8 |
Divide both sides by 8 |
x |
= |
3 |
Now, substitute the value of x into Equation 1
2x+y |
= |
7 |
Equation 1 |
2(3)+y |
= |
7 |
x=3 |
6+y -6 |
= |
7 -6 |
Subtract 6 from both sides |
y |
= |
1 |
Check our work
To check if the values of x and y are correct, substitute both values to the first equation
2x+y |
= |
7 |
2(3)+1 |
= |
7 |
Substitute x=3 and y=1 |
6+1 |
= |
7 |
7 |
= |
7 |
This proves that the values of x and y are correct.
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Question 4 of 5
Solve the following simultaneous equations by substitution.
3x-2y=20
x+2y=-4
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
First, label the two equations 1 and 2 respectively.
3x-2y |
= |
20 |
Equation 1 |
x+2y |
= |
-4 |
Equation 2 |
Next, solve for x in Equation 2.
x+2y |
= |
-4 |
x+2y -2y |
= |
-4 -2y |
Subtract 2y from both sides |
x |
= |
-4-2y |
Simplify |
Substitute x into Equation 1.
3x -2y |
= |
20 |
Equation 1 |
3(-4-2y) -2y |
= |
20 |
x=-4-2y |
-12-6y-2y |
= |
20 |
Distribute 3 inside the parenthesis |
-12-8y |
= |
20 |
Simplify |
-12-8y +12 |
= |
20 +12 |
Solve for y |
-8y |
= |
32 |
-8y ÷(-8) |
= |
32 ÷(-8) |
Divide both sides by -8 |
y |
= |
-4 |
Now, substitute the value of y into Equation 2
x+2y |
= |
-4 |
Equation 2 |
x+2(-2) |
= |
-4 |
y=-2 |
x-8 +8 |
= |
-4 +8 |
Add 8 to both sides |
x |
= |
4 |
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Question 5 of 5
Solve the following simultaneous equations by substitution.
a-b=5
2a+b=4
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
First, label the two equations 1 and 2 respectively.
a-b |
= |
5 |
Equation 1 |
2a+b |
= |
4 |
Equation 2 |
Next, solve for a in Equation 1.
a-b |
= |
5 |
a-b +b |
= |
5 +b |
Add b to both sides |
a |
= |
5+b |
Simplify |
Substitute a into Equation 2.
2a +b |
= |
4 |
Equation 2 |
2(5+b) +b |
= |
4 |
a=5+b |
10+2b+b |
= |
4 |
Distribute 2 inside the parenthesis |
10+3b |
= |
4 |
Simplify |
10+3b -10 |
= |
4 -10 |
Solve for b |
3b |
= |
-6 |
3b ÷3 |
= |
-6 ÷3 |
Divide both sides by 3 |
b |
= |
-2 |
Now, substitute the value of b into Equation 1
a- b |
= |
5 |
Equation 1 |
a- (-2) |
= |
5 |
b=2 |
a+2 |
= |
5 |
Simplify |
a+2 -2 |
= |
5 -2 |
Subtract 2 from both sides |
a |
= |
3 |