Determine the solution sets of the following systems of linear equations.
(I)(II)2y3x==2x−4010−2y
Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
You can solve linear systems of equations using the substitution method, the addition method, or by equating coefficients.
The substitution method is best suited here.
Substitution method
Given is: (I)2y=2x−40(II)3x=10−2y
Equation (I) is solved for 2y. This term also appears in equation (II). So plug equation (I) into (II).
(II) 3x = 10−2y ↓ from eq (I), plug in 2y=2x−40
3x = 10−(2x−40) ↓ multiply out the bracket
3x = 10−(2x−40) +2x ↓ solve for x
5x = 50 :5 x = 10 To find y, plug the value of x into (I).
(I) 2y = 2x−40 ↓ plug in x=10
2y = 2⋅10−40 2y = 20−40 2y = −20 :2 y = −10 Write down the solution set.
L={(x∣y)}={(10∣−10)}
Solution by other methods
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(I)21x−53y=3(II)41x+y=8
Für diese Aufgabe benötigst Du folgendes Grundwissen: System of linear equations
You can solve linear systems of equations using the substitution method, the addition method, or by equating coefficients.
The substitution method is best suited here.
Substitution method
Given is: (I)21x−53y=3(II)41x+y=8
Rewrite (II) such that y is on one side.
(II)41x+y = 8 −41x (II’) y = 8−41x Plug y=8−41x into (I) and solve for x.
(I) 21x−53y = 3 ↓ plug in y=8−41x
21x−53⋅(8−41x) = 3 21x−524+203x = 3 +524 21x+203x = 3+524 ↓ compute the fractions
2010x+203x = 3+4 54 2013x = 7 54 ⋅1320 x = 12 plug x=12 into (II’).
(II’) y = 8−41x ↓ plug in x=12
= 8−41⋅12 = 8−3 = 5 Write down the solution set, first the solution for x, then for y.
L={(12∣5)}
Solution by other methods
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