J 7 m=120 incomePrice = 24 → X, good ILe•P₂ = 3€ →→ X₂ good 23/5U (X₁, X₂) = x₁²/5x₂3-3/S3/5SollutionauMRS = 2x₁quX215X ₂axatheHanget conditionP₁ = 2X₂P₂3X₁we solve the equations³p₁x₁=2p₂X₂P₁ X₁ +P₂X₂ = mthe solutionis:2mSpix₂ = 2mSP₂Substituting prices for income and prices of goods in thisbe the sollution is:how found 24?X₁₂₁ = 24 and 7₁₂=24Please explaine to me step by step thesollution of this exercise so that I canunderstand how the result was foundWDATA-2/52 Xe3x1

For this we already have equation as : 3p1x1=2p2x2 -----(1)and p1x1+p2x2=m------(2)

Answered: m=120 income = 2* →> x₁ good I \p₁₂ =…

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter12: Algebra Of Matrices

Section12.CR: Review Problem Set

Problem 37CR

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