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Question 1 of 2
Solve the following simultaneous equations by substitution.
a−2b+c=−4
2a+2b−c=10
−3a−b−4c=9
Incorrect
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Substitution Method
- 1) Make one variable the subject in one equation
- 2) Substitute this variable in to the other two equations
- 3) Solve these two equations using substitution and find 2 variables
- 4) Substitute the 2 variables into the original equation
First, label the equations 1, 2 and 3 respectively.
a−2b+c |
= |
−4 |
Equation 1 |
2a+2b−c |
= |
10 |
Equation 2 |
−3a−b−4c |
= |
9 |
Equation 3 |
Next, make a the subject in Equation 1.
a−2b+c |
= |
−4 |
a−2b+c +(2b−c) |
= |
−4 +(2b−c) |
Add 2b−c to both sides |
a |
= |
2b−c−4 |
Simplify |
Substitute Equation 1’s a into both Equations 2 and 3. Label the results as Equations A and B
2a+2b−c |
= |
10 |
Equation 2 |
2(2b−c−4)+2b−c |
= |
10 |
a=2b−c−4 |
4b−2c−8+2b−c |
= |
10 |
Distribute 2 inside the parenthesis |
6b−3c−8 |
= |
10 |
Simplify |
6b−3c−8 +8 |
= |
10 +8 |
Add 8 to both sides |
6b−3c |
= |
18 |
Equation A |
−3a−b−4c |
= |
9 |
Equation 3 |
−3(2b−c−4)−b−4c |
= |
9 |
a=2b−c−4 |
−6b+3c+12−b−4c |
= |
9 |
Distribute 2 inside the parenthesis |
−7b−c+12 |
= |
9 |
Simplify |
−7b−c+12 −12 |
= |
9 −12 |
Subtract 12 from both sides |
−7b−c |
= |
−3 |
Equation B |
Next, make c the subject in Equation B.
−7b−c |
= |
−3 |
Equation B |
−7b−c +c |
= |
−3 +c |
Add c to both sides |
−7b |
= |
−3+c |
−7b +3 |
= |
−3+c +3 |
Add 3 to both sides |
−7b+3 |
= |
c |
c |
= |
−7b+3 |
Now, substitute c=−7b+3 into Equation A and solve for b.
6b−3c |
= |
18 |
Equation A |
6b−3(−7b+3) |
= |
18 |
c=−7b+3 |
6b+21b−9 |
= |
18 |
Distribute 3 inside the parenthesis |
27b−9 |
= |
18 |
Simplify |
27b−9 +9 |
= |
18 +9 |
Add 9 to both sides |
27b |
= |
27 |
27b ÷27 |
= |
27 ÷27 |
Divide both sides by 27 |
b |
= |
1 |
Next, substitute the value of b into Equation B and solve for c
c |
= |
−7b +3 |
Equation B |
c |
= |
−7(1)+3 |
b=1 |
c |
= |
−7+3 |
c |
= |
−4 |
Finally, substitute the known values of b and c into Equation 1 to solve for a.
a−2b+c |
= |
−4 |
Equation 1 |
a−2(1)+−4 |
= |
−4 |
b=1 and c=−4 |
a−2−4 |
= |
−4 |
a−6 |
= |
−4 |
Simplify |
a−6 +6 |
= |
−4 +6 |
Add 6 to both sides |
a |
= |
2 |
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Question 2 of 2
Solve the following simultaneous equations by substitution.
3x−3y+z=−2
2y−z=−5
x+4y+2z=−9
Incorrect
Loaded: 0%
Progress: 0%
0:00
Substitution Method
- 1) Make one variable the subject in one equation
- 2) Substitute this variable in to the other two equations
- 3) Solve these two equations using substitution and find 2 variables
- 4) Substitute the 2 variables into the original equation
First, label the equations 1, 2 and 3 respectively.
3x−3y+z |
= |
−2 |
Equation 1 |
2y−z |
= |
−5 |
Equation 2 |
x+4y+2z |
= |
−9 |
Equation 3 |
Next, make z the subject in Equation 1.
2y−z |
= |
−5 |
2y−z +5 |
= |
−5 +5 |
Add 5 to both sides |
2y−z+5 |
= |
0 |
2y+5−z +z |
= |
0 +z |
Add z to both sides |
2y+5 |
= |
z |
z |
= |
2y+5 |
Simplify |
Substitute z=2y+5 into both Equations 1 and 3. Label the results as Equations A and B
3x−3y+z |
= |
−2 |
Equation 1 |
3x−3y+2y+5 |
= |
−2 |
z=2y+5 |
3x−y+5 |
= |
−2 |
Combine like terms |
3x−y+5 −5 |
= |
−2 −5 |
Subtract 5 from both sides |
3x−y |
= |
−7 |
Equation A |
x+4y+2z |
= |
−9 |
Equation 3 |
x+4y+2(2y+5) |
= |
−9 |
z=2y+5 |
x+4y+4y+10 |
= |
−9 |
Distribute 2 inside the parenthesis |
x+8y+10 |
= |
−9 |
Combine like terms |
x+8y+10 −10 |
= |
−9 −10 |
Subtract 10 from both sides |
x+8y |
= |
−19 |
Equation B |
Next, make y the subject in Equation A.
3x−y |
= |
−7 |
Equation A |
3x−y +7 |
= |
−7 +7 |
Add 7 to both sides |
3x+7−y |
= |
0 |
3x+7−y +y |
= |
0 +y |
Add y to both sides |
3x+7 |
= |
y |
y |
= |
3x+7 |
Now, substitute y=3x+7 into Equation B and solve for x.
x+8y |
= |
−19 |
Equation B |
x+8(3x+7) |
= |
−19 |
y=3x+7 |
x+24x+56 |
= |
−19 |
Distribute 8 inside the parenthesis |
25x+56 |
= |
−19 |
Combine like terms |
25x+56 −56 |
= |
−19 −56 |
Subtract 56 from both sides |
25x |
= |
−75 |
27x ÷25 |
= |
−75 ÷25 |
Divide both sides by 25 |
x |
= |
−3 |
Next, substitute the value of x into Equation A and solve for y
3x−y |
= |
−7 |
Equation A |
3(−3)−y |
= |
−7 |
x=−3 |
−9−y |
= |
−7 |
−9−y +9 |
= |
−7 +9 |
Add 9 to both sides |
−y |
= |
2 |
−y ×(−1) |
= |
2 ×(−1) |
Multiply both sides by −1 |
y |
= |
−2 |
Finally, substitute the known value of y into Equation 2 to solve for z.
2y−z |
= |
−5 |
Equation 2 |
2(−2)−z |
= |
−5 |
y=−2 |
−4−z |
= |
−5 |
−4−z +4 |
= |
−5 +4 |
Add 4 to both sides |
−z |
= |
−1 |
−z ×(−1) |
= |
−1 ×(−1) |
Multiply both sides by −1 |
z |
= |
1 |