solve the following by cramers rule x+2y+4=0;3x=-4y-6
Answers
Answer:
Step-by-step explanation:
Example 4. Solve this system of simultaneous equations:
1) 3x + 4y = 19
2) 2x − y = 9
Solution. If we add the equations as they are, neither one of the unknowns will cancel. Now, if the coefficient of y in equation 2) were −4, then the y's would cancel. Therefore we will expand our strategy as follows:
Make one pair of coefficients negatives of one another -- by multiplying
both sides of an equation by the same number. Upon adding the equations, that unknown will be eliminated.
To make the coefficients of the y's 4 and −4, we will multiply both sides of equation 2) by 4 :
1) 3x + 4y = 19 simultaneous equations 3x + 4y = 19
2) 2x − y = 9 simultaneous equations 8x − 4y = 36
simultaneous equations
11x = 55
x = 55
11
x = 5
The 4 over the arrow in equation 2) signifies that both sides of that equation have been multiplied by 4. Equation 1) has not been changed.
To solve for y, substitute x = 5 in either one of the original equations. In equation 1):
3· 5 + 4y = 19
4y = 19 − 15
4y = 4
y = 1
The solution is (5, 1).
The student should always verify the solution by replacing x and y with (5, 1) in the original equations.
Example 5. Solve simultaneously:
1) 3x + 2y = −2
2) 2x + 5y = −5
Solution. We must make one pair of coefficients negatives of one another. In this example, we must decide which of the unknowns to eliminate, x or y. In either case, we will make the new coefficients the Lowest Common Multiple (LCM) of the original coefficients -- but with opposite signs.
Thus, if we eliminate x, then we will make the new coeffients 6 and −6. (The LCM of 3 and 2 is 6.) While if we eliminate y, we will make their new coefficients 10 and −10. (The LCM of 2 and 5 is 10.)
Let us choose to eliminate x:
1) 3x + 2y = −2 simultaneous equations 6x + 4y = −4
2) 2x + 5y = −5 simultaneous equations −6x − 15y = 15
________________________________________________________________________
− 11y = 11
y = −1.
Equation 1) has been multiplied by 2. Equation 2) has been multiplied by −3 -- because we want to make those coefficients 6 and −6, so that on adding, they will cancel.
To solve for x, we will substitute y = −1 in the original equation 1):
3x + 2(−1) = −2
3x − 2 = −2
3x = 0
x = 0
The solution is (0, −1).
Problem 3. Solve simultaneously.
1) 2x + 3y = 13
2) 5x − y = 7
To make the y's cancel, multiply equation 2) by 3:
1) 2x + 3y = 13 simultaneous equations 2x + 3y = 13
2) 5x − y = 7 simultaneous equations 15x − 3y = 21
________________________________________________________________________
17x = 34
x = 2
To solve for y:
Substitute x = 2 in one of the original equations.
In equation 1:
2· 2 + 3y = 13
4 + 3y = 13
3y = 9
y = 3
The solution is (2, 3).
Problem 4. Solve simultaneously.
1) x + 2y = −1
2) 2x − 3y = 5
To make the x's cancel, multiply equation 1) by −2:
1) x + 2y = −1 simultaneous equations −2x − 4y = 2
2) 2x − 3y = 5 simultaneous equations 2x − 3y = 5
________________________________________________________________________
− 7y = 7
y = −1
To solve for x:
Substitute y = −1 in one of the original equations.
In equation 1:
x + 2(−1) = −1
x − 2 = −1
x = −1 + 2
x = 1
The solution is (1, −1).
We could have eliminated y by multiplying equation 1) by 3 and equation 2) by 2.
Problem 5. Solve simultaneously:
1) 3x − 4y = 1
2) 2x + 3y = 12
To make the y's cancel:
Multiply equation 1) by 3 and equation 2) by 4:
1) 3x − 4y = 1 simultaneous equations 9x − 12y = 3
2) 2x + 3y = 12 simultaneous equations 8x + 12y = 48
________________________________________________________________________
17x = 51
x = 51
17
x = 3
To solve for y:
Substitute x = 3 in one of the original equations.
In equation 2 (because the sign of y is already positive):
2· 3 + 3y = 12
6 + 3y = 12
3y = 6
y = 2
The solution is (3, 2).
Problem 6. Solve simultaneously:
1) 3x + 2y = −4
2) 2x + 5y = 1
To make the x's cancel:
Multiply equation 1) by 2 and equation 2) by −3:
1) 3x + 2y = −4 simultaneous equations 6x + 4y = −8
2) 2x + 5y = 1 simultaneous equations −6x − 15y = −3
________________________________________________________________________
− 11y = −11
y = 1
To solve for x:
Substitute y = 1 in one of the original equations.
In equation 1:
3x + 2· 1 = −4
3x + 2 = −4
3x = −4 − 2
3x = −6
x = −2
The solution is (−2, 1).
We could have eliminated y by multiplying equation 1) by 5 and equation 2) by −2.
Problem 7. Solve simultaneously:
1) 5x + 3y = −11
2) 2x + 4y = −10
To make the x's cancel:
Multiply equation 1) by 2 and equation 2) by −5:
1) 5x + 3y = −11 simultaneous equations 10x + 6y = −22
2) 2x + 4y = −10 simultaneous equations −10x − 20y = 50
________________________________________________________________________
− 14y = 28
y = −2
To solve for x:
Substitute y = −2 in one of the original equations.
In equation 1:
5x + 3(−2) = −11
5x − 6 = −11
5x = −11 + 6
5x = −5
x = −1
The solution is (−1, −2).
Answer:
Step-by-step explanation:
