
How do you solve using the elimination method $3x+4y=5$ and $6x+8y=10$ ?
Answer
542.1k+ views
Hint: The process of solving any pair of equations by elimination method which is also known as the combination method is done by eliminating any one of the variables by adding or subtracting both the equations. First, multiply the first linear equation with an integer which on multiplying would result in the constant before the $x$ variable in the second linear equation and subtract them both to cancel one variable and get a solution for the other.
Complete step by step solution:
The given equations are $3x+4y=5$ and $6x+8y=10$
Now consider the first equation, which is $3x+4y=5$
Multiply it with an integer which on multiplying would result in the constant before the $x$ variable in the second equation $6x+8y=10$ which would be $2$
$\Rightarrow 2\left( 3x+4y=5 \right)$
Open the brackets and multiply the constant.
$\Rightarrow 6x+8y=10$
Now subtract both the equations.
$6x+8y=10$
$(-)6x+8y=10$
Now on further simplification, we get,
$\Rightarrow 0$
Hence, we got the answer as zero because the given linear equations are the same. One equation is obtained by multiplying with an integer which is 2.
These equations have an infinite number of solutions.
Note: Any pair of linear equations can be solved by using three methods which are,
The elimination method, the cross-multiplication method, and the substitution method. In the elimination or combination method, we add or subtract both the equations to eliminate one variable and find the solution of others and then substitute and get the solution of the other one also. In the substitution method, we represent one variable in the form of another and substitute it in the other equation to get the solution.
This question can also be solved by the cross-multiplication method which is, we calculate using the formula, $x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}};y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$ .Write the equations in general form and then substitute the values to get the solution for both the equations.
Complete step by step solution:
The given equations are $3x+4y=5$ and $6x+8y=10$
Now consider the first equation, which is $3x+4y=5$
Multiply it with an integer which on multiplying would result in the constant before the $x$ variable in the second equation $6x+8y=10$ which would be $2$
$\Rightarrow 2\left( 3x+4y=5 \right)$
Open the brackets and multiply the constant.
$\Rightarrow 6x+8y=10$
Now subtract both the equations.
$6x+8y=10$
$(-)6x+8y=10$
Now on further simplification, we get,
$\Rightarrow 0$
Hence, we got the answer as zero because the given linear equations are the same. One equation is obtained by multiplying with an integer which is 2.
These equations have an infinite number of solutions.
Note: Any pair of linear equations can be solved by using three methods which are,
The elimination method, the cross-multiplication method, and the substitution method. In the elimination or combination method, we add or subtract both the equations to eliminate one variable and find the solution of others and then substitute and get the solution of the other one also. In the substitution method, we represent one variable in the form of another and substitute it in the other equation to get the solution.
This question can also be solved by the cross-multiplication method which is, we calculate using the formula, $x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}};y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$ .Write the equations in general form and then substitute the values to get the solution for both the equations.
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