How do you find the solution of the system \[2x+4y=8\] and \[x+4y=6\]?
Answer
616.2k+ views
Hint: In the given question, a pair of linear equations consist of two variables which is ‘x’ and ‘y’. Hence this is the question of linear equations in two variables. A pair of linear equations in two variables can be solved using a substitution method. We need to first find the value of one variable in terms of another variable and then substitute that in one of the equations, we will get our required solution set.
Complete step by step solution:
We have given the two equations:
\[2x+4y=8\]------- (1)
\[x+4y=6\]------- (2)
From equation (1), we obtain
\[2x+4y=8\]
\[y=\dfrac{8-2x}{4}\]-------- (3)
Substituting the value of \[y=\dfrac{8-2x}{4}\] in equation (2), we obtain
\[x+4y=6\]
\[x+4\left( \dfrac{8-2x}{4} \right)=6\]
Simplifying the above equation, we get
\[x+8-2x=6\]
Combining the like terms,
\[-x+8=6\]
Subtracting 8 to both the sides of the equation, we get
\[-x+8-8=6-8\]
Simplifying the above equation, we get
\[-x=-2\]
Divide both the side of equation by -1, we get
\[x=2\]
Therefore,
\[\Rightarrow x=2\]
Substitute the value of \[x=2\] in \[y=\dfrac{8-2x}{4}\],
\[y=\dfrac{8-2\left( 2 \right)}{4}=\dfrac{8-4}{4}=\dfrac{4}{4}=1\]
Therefore,
\[\Rightarrow y=1\]
Therefore, a pair of linear equations in two variables has a solution set of (x, y) = \[\left( 2,1 \right)\].
Note: A pair of linear equations in two variables have two solutions, one solution for the ‘x’ i.e. the value of ‘x’ and other solution for the ‘y’ i.e. the value of ‘y’. The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. The solution of the equation will not change if the same number is added to or subtracted from both the sides of the equation, multiplying and dividing both the sides of the equations by the same non-zero number.
Complete step by step solution:
We have given the two equations:
\[2x+4y=8\]------- (1)
\[x+4y=6\]------- (2)
From equation (1), we obtain
\[2x+4y=8\]
\[y=\dfrac{8-2x}{4}\]-------- (3)
Substituting the value of \[y=\dfrac{8-2x}{4}\] in equation (2), we obtain
\[x+4y=6\]
\[x+4\left( \dfrac{8-2x}{4} \right)=6\]
Simplifying the above equation, we get
\[x+8-2x=6\]
Combining the like terms,
\[-x+8=6\]
Subtracting 8 to both the sides of the equation, we get
\[-x+8-8=6-8\]
Simplifying the above equation, we get
\[-x=-2\]
Divide both the side of equation by -1, we get
\[x=2\]
Therefore,
\[\Rightarrow x=2\]
Substitute the value of \[x=2\] in \[y=\dfrac{8-2x}{4}\],
\[y=\dfrac{8-2\left( 2 \right)}{4}=\dfrac{8-4}{4}=\dfrac{4}{4}=1\]
Therefore,
\[\Rightarrow y=1\]
Therefore, a pair of linear equations in two variables has a solution set of (x, y) = \[\left( 2,1 \right)\].
Note: A pair of linear equations in two variables have two solutions, one solution for the ‘x’ i.e. the value of ‘x’ and other solution for the ‘y’ i.e. the value of ‘y’. The important thing to recollect about any equation is that the ‘equals’ sign represents a balance. What the sign says is that what’s on the left-hand side is strictly an equal to what’s on the right-hand side. The solution of the equation will not change if the same number is added to or subtracted from both the sides of the equation, multiplying and dividing both the sides of the equations by the same non-zero number.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

