solve 5x+y=20, 10x-2y=50 by substitution method pleas Koi solve kr do important About the author Ruby
Basic Concept Used :- To solve systems using substitution, follow this procedure: Select one equation and solve it to get one variable in terms of second variables. In the second equation, substitute the value of variable evaluated in Step 1 to reduce the equation to one variable. Solve the new equation to get the value of one variable. Substitute the value found in to any one of two equations involving both variables and solve for the other variable. Let’s solve the problem now!! ↝ Given Linear Equations are [tex]\rm :\longmapsto\:5x + y = 20 – – – (1)[/tex] and [tex]\rm :\longmapsto\:10x – 2y = 50 – – – (2)[/tex] ↝ From equation (1), we have [tex]\rm :\longmapsto\:y \: = \: 20 – 5x – – – (3)[/tex] ↝ Substituting the value of y in equation (2), [tex]\rm :\longmapsto\:10x – 2(20 – 5x) = 50[/tex] [tex]\rm :\longmapsto\:10x – 40 + 10x = 50[/tex] [tex]\rm :\longmapsto\:20x = 50 + 40[/tex] [tex]\rm :\longmapsto\:20x = 90[/tex] [tex]\bf\implies \:x = \dfrac{9}{2} – – – (4)[/tex] ↝ On substituting the value of x in equation (3), we get [tex]\rm :\longmapsto\:y = 20 – 5 \times \dfrac{9}{2} [/tex] [tex]\rm :\longmapsto\:y = 20 – \dfrac{45}{2} [/tex] [tex]\rm :\longmapsto\:y = \dfrac{40 – 45}{2} [/tex] [tex]\bf\implies \:y \: = \: – \: \dfrac{5}{2} [/tex] [tex]\overbrace{ \underline { \boxed { \bf \therefore The \: solution\: is \: x \: = \: \dfrac{9}{2} \: \: and \: \:y \: = – \: \dfrac{5}{2}}}}[/tex] Additional Information There are 4 methods to solve this type of pair of linear equations. 1. Method of Substitution 2. Method of Eliminations 3. Method of Cross Multiplication 4. Graphical Method Reply
Answer: x = 9/2 and y = -5/2 Step-by-step explanation: The given equations are : 5x + y = 20 …(1) 10x – 2y = 50 …(2) From equation (1), we get; 5x + y = 20 5x = 20 -y x = (20 – y)/5 Substituting x = (20 – y)/5 in equation (2), we get; 10 (20 – y)/5 – 2y = 50 (200 – 10y)/5 -2y = 50 (200 – 10y -10y)/5 = 50 200 – 20y = 50 x 5 200 – 20y = 250 -20y = 250 – 200 -20y = 50 y = 50/-20 y = -5/2 Substituting the value of y in equation (1), we get; 5x + y = 20 5x + (-5/2) = 20 5x – 5/2 = 20 (10x – 5)/2 = 20 10x – 5 = 20 x 2 10x – 5 = 40 10x = 40 + 5 10x = 45 x = 45/10 x = 9/2 Hope it helps u… Reply
Basic Concept Used :-
To solve systems using substitution, follow this procedure:
Let’s solve the problem now!!
↝ Given Linear Equations are
[tex]\rm :\longmapsto\:5x + y = 20 – – – (1)[/tex]
and
[tex]\rm :\longmapsto\:10x – 2y = 50 – – – (2)[/tex]
↝ From equation (1), we have
[tex]\rm :\longmapsto\:y \: = \: 20 – 5x – – – (3)[/tex]
↝ Substituting the value of y in equation (2),
[tex]\rm :\longmapsto\:10x – 2(20 – 5x) = 50[/tex]
[tex]\rm :\longmapsto\:10x – 40 + 10x = 50[/tex]
[tex]\rm :\longmapsto\:20x = 50 + 40[/tex]
[tex]\rm :\longmapsto\:20x = 90[/tex]
[tex]\bf\implies \:x = \dfrac{9}{2} – – – (4)[/tex]
↝ On substituting the value of x in equation (3), we get
[tex]\rm :\longmapsto\:y = 20 – 5 \times \dfrac{9}{2} [/tex]
[tex]\rm :\longmapsto\:y = 20 – \dfrac{45}{2} [/tex]
[tex]\rm :\longmapsto\:y = \dfrac{40 – 45}{2} [/tex]
[tex]\bf\implies \:y \: = \: – \: \dfrac{5}{2} [/tex]
[tex]\overbrace{ \underline { \boxed { \bf \therefore The \: solution\: is \: x \: = \: \dfrac{9}{2} \: \: and \: \:y \: = – \: \dfrac{5}{2}}}}[/tex]
Additional Information
There are 4 methods to solve this type of pair of linear equations.
Answer:
x = 9/2 and y = -5/2
Step-by-step explanation:
The given equations are :
5x + y = 20 …(1)
10x – 2y = 50 …(2)
From equation (1), we get;
5x + y = 20
5x = 20 -y
x = (20 – y)/5
Substituting x = (20 – y)/5 in equation (2), we get;
10 (20 – y)/5 – 2y = 50
(200 – 10y)/5 -2y = 50
(200 – 10y -10y)/5 = 50
200 – 20y = 50 x 5
200 – 20y = 250
-20y = 250 – 200
-20y = 50
y = 50/-20
y = -5/2
Substituting the value of y in equation (1), we get;
5x + y = 20
5x + (-5/2) = 20
5x – 5/2 = 20
(10x – 5)/2 = 20
10x – 5 = 20 x 2
10x – 5 = 40
10x = 40 + 5
10x = 45
x = 45/10
x = 9/2
Hope it helps u…