Can anybody answer this, please? you'll get 27 points.
2 answers:
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Answer:
(x, y) = (4, -3)
Step-by-step explanation:
- x-3y=13
- 2x+4y=-4
First of all, look at the given equations. Here, we see that the second equation has coefficients that all have a factor of 2. If we divide that out, we get an equation that has an x-coefficient of 1, matching the x-coefficient in the first equation.
Here is the reduced second equation:
x +2y = -2
Now, if we subtract one equation from the other, the variable x will be eliminated. We want to choose that subtraction wisely.
We note that the y-coefficient in the first equation is less than that in the second equation. If we subtract the first equation from the second, the result will have a positive y-coefficient:
(x +2y) -(x -3y) = (-2) -(13)
5y = -15 . . . . . . . . . . . . . . . . simplify. Note that the x-variable is eliminated, which is the purpose of this exercise.
y = -3 . . . . . . . divide by 5
We can use this value in the reduced second equation to find the value of x:
x +2(-3) = -2
x = 4 . . . . . . . . . add 6 and simplify
The solution is (x, y) = (4, -3).
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I find a graphing calculator provides an easy and reliable check of the answer.
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<em>Comment on linear combination</em>
This method is often called "elimination," because the purpose of combining the equations in a particular way is to eliminate one of the variables. This requires you look at the coefficients of the variables and devise a plan to combine them so the resulting coefficient for one of the variables is zero.
In the worst case, you can combine ...
- ax +by = c
- dx +ey = f
by multiplying the second equation by <em>a</em> and the first by <em>-d</em>:
a(dx +ey) -d(ax +by) = a( f) -d(c)
y(ae -bd) = fa -cd . . . . . . simplify; x is eliminated
This sort of approach results in a formula for the solution known as Cramer's Rule.
y = (fa-cd)/(ae-bd)
The corresponding solution for x is ...
x = (ce-bf)/(ae-bd)
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The point of looking at the equations first is that you can often choose which variable to eliminate and what multiplier to use to minimize the amount of arithmetic involved—as we did above.
