if we multiply 81 divide both sides if linear equation by nm - zero number, their the mts of linear equation will remain the same, is it true? if so justify with an example
Answers
Step-by-step explanation:
Let us take an equation x+2y=5 (1)
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHS
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHS
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.Let us divide both sides of equation (1) by 2
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.Let us divide both sides of equation (1) by 2⇒2x+y=25
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.Let us divide both sides of equation (1) by 2⇒2x+y=25Put x=1,y=2 in above equation
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.Let us divide both sides of equation (1) by 2⇒2x+y=25Put x=1,y=2 in above equationLHS =21+2=25= RHS
Let us take an equation x+2y=5 (1)Put x=1 in eqn (1)⇒y=2So,x=1,y=2 is a solution of the equation Multiplying both sides of equation 1() by 3, we get3x+6y=15Put x=1,y=2 in LHSLHS =3×1+6×2=15= RHSHence, on multiplying an equation by a non-zero number , the solution remains same.Let us divide both sides of equation (1) by 2⇒2x+y=25Put x=1,y=2 in above equationLHS =21+2=25= RHSHence, on dividing or multiplying an equation by a non-zero number, the solution remains the same.
Answer:
the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a by-product.the process by which green plants and some other organisms use sunlight to synthesize nutrients from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a by-product.