
How do you solve the following linear system: \[3x + 5y = - 1\] , \[2x - 5y = 16\]?
Answer
536.1k+ views
Hint: Given equations are standard form of \[Ax + By = C\]. To solve the given simultaneous linear equation, combine all the terms by using any of the elementary arithmetic functions i.e., addition, subtraction, multiplication and division, then to get equation for x or y then simplify the terms to get the value of \[x\] also the value of \[y\].
Complete step by step solution:
Given,
\[3x + 5y = - 1\] ……………………….1
\[2x - 5y = 16\] ………………………. 2
The standard form of equation is
\[Ax + By = C\]
Starting with the first equation, we need to make either x or y the subject, let us solve with respect to x as:
\[3x + 5y = - 1\]
Shift the terms, to solve for x we get:
\[ \Rightarrow 3x = - 5y - 1\]
Hence, we get the equation for x as:
\[ \Rightarrow x = \dfrac{{ - 5y - 1}}{3}\] ………………. 3
Now, substitute value of x i.e., from equation 3 in equation 2 as:
\[2x - 5y = 16\]
\[ \Rightarrow 2\left( {\dfrac{{ - 5y - 1}}{3}} \right) - 5y = 16\]
Multiplying the terms, we get:
\[ \Rightarrow \dfrac{{ - 10y - 2}}{3} - 5y = 16\]
\[ \Rightarrow - 10y - 2 - 5y\left( 3 \right) = 16\left( 3 \right)\]
Evaluating the terms, we get:
\[ \Rightarrow - 10y - 2 - 15y = 48\]
\[ \Rightarrow - 25y = 48 + 2\]
\[ \Rightarrow - 25y = 50\]
\[ \Rightarrow y = - \dfrac{{50}}{{25}}\]
Hence, the value of y is:
\[ \Rightarrow y = - 2\]
Now, substitute y=-2 in equation 2, we get:
\[2x - 5y = 16\]
\[ \Rightarrow 2x - 5\left( { - 2} \right) = 16\]
\[ \Rightarrow 2x + 10 = 16\]
Evaluating the terms, we get:
\[ \Rightarrow 2x = 16 - 10\]
\[ \Rightarrow 2x = 6\]
\[ \Rightarrow x = \dfrac{6}{2}\]
Hence, the value of x is:
\[ \Rightarrow x = 3\]
Now, let us verify the answers, which we have obtained in the given equation as:
\[3x + 5y = - 1\]
Substitute value of x and y as, \[x = 3\] and \[y = - 2\] in equation 1 we get:
\[ \Rightarrow 3\left( 3 \right) + 5\left( { - 2} \right) = - 1\]
\[ \Rightarrow 9 - 10 = - 1\]
\[ \Rightarrow - 1 = - 1\]
As, we got LHS=RHS this means that our obtained values are correct for the given equations.
Therefore, the values of x and y are:
\[x = 3\] and \[y = - 2\].
Note: The main key point to solve the equations is that we need to find an equation for either x or y, then substitute that value in the other equation. As we know that Simultaneous equations are two equations, each with the same two unknowns and are simultaneous because they are solved together, hence the key point to solve this kind of equations is we need to combine all the terms and then simplify the terms to get the value of \[x\] also the value of \[y\].
Complete step by step solution:
Given,
\[3x + 5y = - 1\] ……………………….1
\[2x - 5y = 16\] ………………………. 2
The standard form of equation is
\[Ax + By = C\]
Starting with the first equation, we need to make either x or y the subject, let us solve with respect to x as:
\[3x + 5y = - 1\]
Shift the terms, to solve for x we get:
\[ \Rightarrow 3x = - 5y - 1\]
Hence, we get the equation for x as:
\[ \Rightarrow x = \dfrac{{ - 5y - 1}}{3}\] ………………. 3
Now, substitute value of x i.e., from equation 3 in equation 2 as:
\[2x - 5y = 16\]
\[ \Rightarrow 2\left( {\dfrac{{ - 5y - 1}}{3}} \right) - 5y = 16\]
Multiplying the terms, we get:
\[ \Rightarrow \dfrac{{ - 10y - 2}}{3} - 5y = 16\]
\[ \Rightarrow - 10y - 2 - 5y\left( 3 \right) = 16\left( 3 \right)\]
Evaluating the terms, we get:
\[ \Rightarrow - 10y - 2 - 15y = 48\]
\[ \Rightarrow - 25y = 48 + 2\]
\[ \Rightarrow - 25y = 50\]
\[ \Rightarrow y = - \dfrac{{50}}{{25}}\]
Hence, the value of y is:
\[ \Rightarrow y = - 2\]
Now, substitute y=-2 in equation 2, we get:
\[2x - 5y = 16\]
\[ \Rightarrow 2x - 5\left( { - 2} \right) = 16\]
\[ \Rightarrow 2x + 10 = 16\]
Evaluating the terms, we get:
\[ \Rightarrow 2x = 16 - 10\]
\[ \Rightarrow 2x = 6\]
\[ \Rightarrow x = \dfrac{6}{2}\]
Hence, the value of x is:
\[ \Rightarrow x = 3\]
Now, let us verify the answers, which we have obtained in the given equation as:
\[3x + 5y = - 1\]
Substitute value of x and y as, \[x = 3\] and \[y = - 2\] in equation 1 we get:
\[ \Rightarrow 3\left( 3 \right) + 5\left( { - 2} \right) = - 1\]
\[ \Rightarrow 9 - 10 = - 1\]
\[ \Rightarrow - 1 = - 1\]
As, we got LHS=RHS this means that our obtained values are correct for the given equations.
Therefore, the values of x and y are:
\[x = 3\] and \[y = - 2\].
Note: The main key point to solve the equations is that we need to find an equation for either x or y, then substitute that value in the other equation. As we know that Simultaneous equations are two equations, each with the same two unknowns and are simultaneous because they are solved together, hence the key point to solve this kind of equations is we need to combine all the terms and then simplify the terms to get the value of \[x\] also the value of \[y\].
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