
Solve the following sets of simultaneous equations.
\[2x + 3y + 4 = 0; x - 5y = 11\]
Answer
521.1k+ views
Hint: Here the given question is to solve for the variables, in the equation. To solve these equations we first need to eliminate one of the variables and then find the value of the other variable, and then after putting the value of the obtained variable we can have the value of the eliminated variable.
Complete step-by-step solution:
Here in the first equation, variable “x” is multiplied by two and in the second equation the variable “x” has the coefficient of one, so here we will multiply the second equation with two, and then subtracts both the equations, on solving we get:
\[\Rightarrow 2x + 3y = - 4 \to eq1 \\
\Rightarrow x - 5y = 11 \to eq2 \]
Multiplying by two in equation second we get:
\[ \Rightarrow 2(x - 5y = 11) \to eq2 \\
\Rightarrow 2x - 10y = 22 \to eq3 \]
Solving equation second and third by subtraction we get:
\[\Rightarrow (2x + 3y) - (2x - 10y) = ( - 4) - (22) \\
\Rightarrow 2x + 3y - 2x + 10y = - 26 \\
\Rightarrow 13y = - 26 \\
\Rightarrow y = \dfrac{{ - 26}}{{13}} = - 2 \]
Now putting the value of “y” in equation one we get:
\[ \Rightarrow 2x + 3( - 2) = - 4 \\
\Rightarrow 2x = - 4 + 6 \\
\Rightarrow x = \dfrac{2}{2} = 1 \]
Here we get the values of both the variables.
Note: To solve the system of equations we can go through the substitution method, in which we need to find one variable in terms of another, then by putting this variable in the remaining equation and on further solving that equation we get the value of a variable.
Complete step-by-step solution:
Here in the first equation, variable “x” is multiplied by two and in the second equation the variable “x” has the coefficient of one, so here we will multiply the second equation with two, and then subtracts both the equations, on solving we get:
\[\Rightarrow 2x + 3y = - 4 \to eq1 \\
\Rightarrow x - 5y = 11 \to eq2 \]
Multiplying by two in equation second we get:
\[ \Rightarrow 2(x - 5y = 11) \to eq2 \\
\Rightarrow 2x - 10y = 22 \to eq3 \]
Solving equation second and third by subtraction we get:
\[\Rightarrow (2x + 3y) - (2x - 10y) = ( - 4) - (22) \\
\Rightarrow 2x + 3y - 2x + 10y = - 26 \\
\Rightarrow 13y = - 26 \\
\Rightarrow y = \dfrac{{ - 26}}{{13}} = - 2 \]
Now putting the value of “y” in equation one we get:
\[ \Rightarrow 2x + 3( - 2) = - 4 \\
\Rightarrow 2x = - 4 + 6 \\
\Rightarrow x = \dfrac{2}{2} = 1 \]
Here we get the values of both the variables.
Note: To solve the system of equations we can go through the substitution method, in which we need to find one variable in terms of another, then by putting this variable in the remaining equation and on further solving that equation we get the value of a variable.
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