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Perplex
  • Dashboard
Topics
Exponents & LogarithmsRounding & ErrorSequences & SeriesCounting & BinomialsProof and ReasoningComplex NumbersAlgebra Skills
Cartesian plane & linesQuadraticsFunction TheoryTransformations & asymptotesPolynomials
2D & 3D GeometryTrig equations & identitiesVectors
ProbabilityDescriptive StatisticsBivariate StatisticsDistributions & Random Variables
DifferentiationIntegrationDifferential EquationsMaclaurin
Paper 3
Plus
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Perplex

Algebra Skills (Lesson 2/2)

3 by 3 Systems of equations

1 / 6

Systems of equations with 2 unknowns

We can solve a system of  ​2​ equations and ​2​ unknowns with different methods.

​
{3y+2x−2=03y−3x+1=0​
​


By substitution

Rearranging

​
3y+2x−2=0⇒y=−32​x+32​
​


Substituting this into ​3y−3x+1=0:

​
(−2x+2)−3x+1=0
​
​
−5x=−3
​

So ​x=53​, which implies ​y=−32​⋅53​+32​=154​. So the intersection is ​(53​,154​).


By elimination

We can eliminate ​y​ from the equations by subtracting the second from the first:

​
(3y+2x−2)−(2y−3x+1)2x−2+3x−15x​=0=0=3​
​

So ​x=53​⇒y=154​​ and the intersection is again ​(53​,154​).


We can use either of these methods to systems of equations with 2 equations and 2 unknowns.

Algebra Skills (Lesson 2/2)

3 by 3 Systems of equations

1 / 6

Systems of equations with 2 unknowns

We can solve a system of  ​2​ equations and ​2​ unknowns with different methods.

​
{3y+2x−2=03y−3x+1=0​
​


By substitution

Rearranging

​
3y+2x−2=0⇒y=−32​x+32​
​


Substituting this into ​3y−3x+1=0:

​
(−2x+2)−3x+1=0
​
​
−5x=−3
​

So ​x=53​, which implies ​y=−32​⋅53​+32​=154​. So the intersection is ​(53​,154​).


By elimination

We can eliminate ​y​ from the equations by subtracting the second from the first:

​
(3y+2x−2)−(2y−3x+1)2x−2+3x−15x​=0=0=3​
​

So ​x=53​⇒y=154​​ and the intersection is again ​(53​,154​).


We can use either of these methods to systems of equations with 2 equations and 2 unknowns.