To solve the simultaneous equations by deterninant method, write D
y + 2x = 10; 2x - 3y = 0
Answers
Answer:
Step-by-step explanation:
olve 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
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− 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
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17x = 34
x = 2
To solve for y:
Substitute x = 2 in one of the original equations.
In equation 1: