The following steps will be useful to solve system of linear equations using method of substitution.
Step 1 :
In the given two equations, solve one of the equations either for x or y.
Step 2 :
Substitute the result of step 1 into other equation and solve for the second variable.
Step 3 :
Using the result of step 2 and step 1, solve for the first variable.
How to solve linear equations by elimination method?
Step 1 :
By taking anyone equations from the given two equations ,first multiply by some suitable non-zero constant to make the co-efficient of one variable (either x or y) numerically equal.
Step 2 :
If both coefficients are numerically equal to the same sign, then we may eliminate them by subtracting those equations.
If they have different signs, then we may add both equations and eliminate them.
Step 3 :
After eliminating one variable, we may get the value of one variable.
Step 4 :
The remaining variable is then found by substituting in any one of the given equations.
Problem 1 :
x = - 2y
x - y = 9
Solution :
x = - 2y ---> (1)
x - y = 9 ---> (2)
Substitute x = -2y in equation (2)
-2y - y = 9
-3y = 9
y = 9/-3
y = -3
By applying y = -3 in equation (1), we get
x = -2(-3)
x = 6
Therefore, the solution is x = 6 and y = -3.
Problem 2 :
3x + y = 7
-2x - y = 9
Solution :
3x + y = 7 ---> (1)
-2x - y = 9 ---> (2)
Add (1) + (2), we get
3x + y - 2x - y = 7 + 9
x = 16
By applying x = 16 in (1) equation, we get
3(16) + y = 7
48 + y = 7
y = 7 - 48
y = -41
Therefore, the solution is x = 16 and y = -41.
Solve each system by substitution.
Problem 1 :
y = -3x + 4 and y = 4x - 10 Solution
Problem 2 :
y = -4x + 2 and y = 6x - 8 Solution
Problem 3 :
y = 2x and -6x + 3y = 16 Solution
Problem 4 :
y = -3x and 4x - 2y = -20 Solution
Problem 5 :
y = 3x - 4 and 4x + 3y = 1 Solution
Problem 6 :
y = x - 4 and -4x - 6y = -16 Solution
Problem 7 :
x = 3y + 1 and 2x + 4y = 12 Solution
Problem 8 :
x = y - 4 and -2x + 3y = 6 Solution
Problem 9 :
Next week your math teacher is giving a chapter test. The test will consist of 35 questions. Some problems are worth 2 points and some problems are worth 4 points. There are 20 questions worth 2 points. How many problems of 4 points are on the test?
Problem 10 :
Next week your math teacher is giving a chapter test. The test will consist of 35 questions. Some problems are worth 2 points and some problems are worth 4 points. There are 20 questions worth 2 points. How many problems of 4 points are on the test?
1) x = 2 and y = -2
2) x = 1 and y = -2.
3) x = 6 and y = -3.
4) no solution.
5) x = -2 and y = 6.
6) x = 1 and y = -1.
7) x = 4 and y = 0.
8) x = 4 and y = 1.
9) x = -6 and y = -2.
10) Therefore 20 questions are worth 2 points and 15 questions are worth 4 points.
Problem 1 :
3x + y = 7
-2x - y = 9
Problem 2 :
4x + 3y = 2
2x - 3y = 1
Problem 3 :
2x + 2y = 3
x = 4y - 1
Problem 4 :
y = 1 + x
2x + y = -2
Problem 5 :
1/2x + 4y = 4
2x - y = 1
Problem 6 :
y = x - 4
4x + y = 26
Problem 7 :
3x + 4y = -23
2y - x = -19
Waht is the solution (x, y) to the system of eqautions above ?
a) (-5, -2) b) (3, -8) c) (4, -6) d) (9, -6)
Problem 8 :
b = 2.35 + 0.25x
c = 1.75 + 0.40x
In the equation above b and c represent the price per pound in dollars, of beef and chicken respectively, x weeks after July 1 during last summer. What was the price per pound of beef when it was equal to the price per pound of chicken ?
1) x = 16 and y = -41
2) x = 1/2 and y = 0
3) x = 1 and y = 1/2
4) x = -1 and y = 0
5) x = 16/17 and y = 15/17
6) x = 6 and y = 2.
7) (3, -8). Option b is correct.
8) $3.35
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM