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Sagot :
Answer:
a = 3
b = 4
Step-by-step explanation:
Simultaneous equation
3a = 2b + 1
3a - 2b = 1 ------------------(I)
3b = 5a - 3
-5a + 3b = -3 --------------(II)
Multiply equation (I) by 3 & equation (II) by 2 and then add. So, now 'b' will be eliminated and we can find the value of 'a'.
(I)*3 9a - 6b = 3
(II)*2 -10a + 6b = -6 {Now add}
-a = -3
a = -3/(-1)
[tex]\sf \boxed{a = 3}[/tex]
Plugin a = 3 in equation (I)
3*3 - 2b = 1
9 - 2b = 1
-2b = 1 - 9 {Subtracting 9 from both sides}
-2b = - 8
b = (-8)/(-2) {Dividing both sides by (-2_}
[tex]\sf \boxed{b = 4}[/tex]
Answer:
(a, b) = (3, 4)
Step-by-step explanation:
In order to eliminate one of the variables in this system of equations, both equations need to be multiplied by different numbers before they are added or subtracted. We can eliminate the 'b' variable by subtracting 3 times the first equation from 2 times the second. (Putting the 'b' terms on the same side of the equal sign makes this easier.)
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2(3b) -3(2b +1) = 2(5a -3) -3(3a)
6b -6b -3 = 10a -6 -9a
3 = a . . . . . . . add 6 and collect terms
3b = 5(3) -3 . . . . use the second equation to find b from a
b = 4 . . . . . . . divide by 3
The solution is (a, b) = (3, 4).
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The multiplier used in each case is the coefficient of the variable in the other equation. The two equations have 'b' coefficients of 2 and 3, so we multiplied the equations by 3 and 2 before subtracting. That makes the 6b terms cancel, leaving only terms with the 'a' variable.
By swapping sides of the equation the way we did, the 'a' terms end up on the same side of the equal sign, which is where we want them. Likewise, the 'b' terms end up on the same side of the equal sign, so their cancellation is easier to see.
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