HomeMathematicsPair of Linear Equations in Two VariablesNCERT SolutionsNCERT ExamplesExample 4Mathematics · Pair of Linear Equations in Two Variables · NCERT ExamplesExample 4A complete, independently verified solution for NCERT Example 4.ChapterSelect a chapterChapter 1: Real NumbersChapter 2: PolynomialsChapter 3: Pair of Linear Equations in Two VariablesChapter 4: Quadratic EquationsChapter 5: Arithmetic ProgressionsChapter 6: TrianglesChapter 7: Coordinate GeometryChapter 8: Introduction to TrigonometryChapter 9: Some Applications of TrigonometryChapter 10: CirclesChapter 11: Areas Related to CirclesChapter 12: Surface Areas and VolumesChapter 13: StatisticsChapter 14: ProbabilityResourceOverviewNCERT ExamplesNCERT ExercisesTextbookPrevious Year QuestionsQuestion BankHOTSModulesExercise / ExampleSelect an entryExample 1Example 2Example 3Example 4Example 5Example 6Example 7Example 8Example 9Example 10Change resourceExample 4ChapterSelect a chapterChapter 1: Real NumbersChapter 2: PolynomialsChapter 3: Pair of Linear Equations in Two VariablesChapter 4: Quadratic EquationsChapter 5: Arithmetic ProgressionsChapter 6: TrianglesChapter 7: Coordinate GeometryChapter 8: Introduction to TrigonometryChapter 9: Some Applications of TrigonometryChapter 10: CirclesChapter 11: Areas Related to CirclesChapter 12: Surface Areas and VolumesChapter 13: StatisticsChapter 14: ProbabilityResourceOverviewNCERT ExamplesNCERT ExercisesTextbookPrevious Year QuestionsQuestion BankHOTSModulesExercise / ExampleSelect an entryExample 1Example 2Example 3Example 4Example 5Example 6Example 7Example 8Example 9Example 10Mathematics · Chapter 3 · NCERT ExamplesPrintDownload PDF← Previous ExampleExample 3Next ExampleExample 5 →Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.NCERT ExampleExample 4substitution methodSolve 7x − 15y = 2 and x + 2y = 3 by substitution.Solutionx=3−2yx=3-2yx=3−2y7(3−2y)−15y=2⇒−29y=−197(3-2y)-15y=2\Rightarrow-29y=-197(3−2y)−15y=2⇒−29y=−19y=1929,x=3−3829=4929y=\frac{19}{29},\qquad x=3-\frac{38}{29}=\frac{49}{29}y=2919,x=3−2938=29497(4929)−15(1929)=2,4929+2(1929)=37\left(\frac{49}{29}\right)-15\left(\frac{19}{29}\right)=2,\quad\frac{49}{29}+2\left(\frac{19}{29}\right)=37(2949)−15(2919)=2,2949+2(2919)=3Final answerx=4929,y=1929x=\frac{49}{29},\qquad y=\frac{19}{29}x=2949,y=2919← Previous ExampleExample 3Next ExampleExample 5 →
NCERT ExampleExample 4substitution methodSolve 7x − 15y = 2 and x + 2y = 3 by substitution.Solutionx=3−2yx=3-2yx=3−2y7(3−2y)−15y=2⇒−29y=−197(3-2y)-15y=2\Rightarrow-29y=-197(3−2y)−15y=2⇒−29y=−19y=1929,x=3−3829=4929y=\frac{19}{29},\qquad x=3-\frac{38}{29}=\frac{49}{29}y=2919,x=3−2938=29497(4929)−15(1929)=2,4929+2(1929)=37\left(\frac{49}{29}\right)-15\left(\frac{19}{29}\right)=2,\quad\frac{49}{29}+2\left(\frac{19}{29}\right)=37(2949)−15(2919)=2,2949+2(2919)=3Final answerx=4929,y=1929x=\frac{49}{29},\qquad y=\frac{19}{29}x=2949,y=2919
