How do you solve the simultaneous equations \[{{x}^{2}}+{{y}^{2}}=5\] and \[y=3x+1\]?
Answer
623.1k+ views
Hint: To solve simultaneous equation in two variables, we need to follow some steps as follows,
Step 1: use one of the equations to find the relationship between the two variables.
Step 2: substitute this relationship in the second equation, to get a one-variable equation.
Step 3: solve this one variable equation, to find the solution values.
Step use any of the equations to find the values of the other variable for each value in step 3.
We will follow these steps for these equations also.
Complete step-by-step answer:
We are given two equations \[{{x}^{2}}+{{y}^{2}}=5\], and \[y=3x+1\].
Substituting \[y=3x+1\] in the equation \[{{x}^{2}}+{{y}^{2}}=5\], we get
\[\Rightarrow {{x}^{2}}+{{\left( 3x+1 \right)}^{2}}=5\]
Simplifying the above equation, we get
\[\Rightarrow 10{{x}^{2}}+6x+1=5\]
\[\Rightarrow 10{{x}^{2}}+6x+-4=0\]
Dividing both sides of the above equation by 2, we get
\[\Rightarrow 5{{x}^{2}}+3x+-2=0\]
We can find the one root of the equation from the hit and trial method, substitute \[x=-1\] in LHS of the above equation, we get
\[\Rightarrow 5{{(-1)}^{2}}+3(-1)+-2=0\]
Hence, one root of the equation is \[-1\]. Using the product of roots property, we get the other root of the equation as \[x=\dfrac{2}{5}\]. Thus, we have to value the variable \[x\], \[-1\] and \[\dfrac{2}{5}\]. Using the relationship between the variables, that is \[y=3x+1\] we can find the values of the variable \[y\] as follows,
For \[x=-1\], \[y=3(-1)+1=-2\].
For \[x=\dfrac{2}{5}\], \[y=3\left( \dfrac{2}{5} \right)+1=\dfrac{11}{5}\].
Hence the solution of the given simultaneous equation are \[x=-1,y=-2\] and \[x=\dfrac{2}{5},y=\dfrac{11}{5}\].
Note: Using the linear equation \[y=3x+1\], we get \[x=\dfrac{y-1}{3}\]. We can also substitute this relationship in the equation \[{{x}^{2}}+{{y}^{2}}=5\]. By this we will get an equation in terms of \[y\] by solving the equation, we will get the solution values of \[y\]. And then we can find the value of x for each \[y\].
Step 1: use one of the equations to find the relationship between the two variables.
Step 2: substitute this relationship in the second equation, to get a one-variable equation.
Step 3: solve this one variable equation, to find the solution values.
Step use any of the equations to find the values of the other variable for each value in step 3.
We will follow these steps for these equations also.
Complete step-by-step answer:
We are given two equations \[{{x}^{2}}+{{y}^{2}}=5\], and \[y=3x+1\].
Substituting \[y=3x+1\] in the equation \[{{x}^{2}}+{{y}^{2}}=5\], we get
\[\Rightarrow {{x}^{2}}+{{\left( 3x+1 \right)}^{2}}=5\]
Simplifying the above equation, we get
\[\Rightarrow 10{{x}^{2}}+6x+1=5\]
\[\Rightarrow 10{{x}^{2}}+6x+-4=0\]
Dividing both sides of the above equation by 2, we get
\[\Rightarrow 5{{x}^{2}}+3x+-2=0\]
We can find the one root of the equation from the hit and trial method, substitute \[x=-1\] in LHS of the above equation, we get
\[\Rightarrow 5{{(-1)}^{2}}+3(-1)+-2=0\]
Hence, one root of the equation is \[-1\]. Using the product of roots property, we get the other root of the equation as \[x=\dfrac{2}{5}\]. Thus, we have to value the variable \[x\], \[-1\] and \[\dfrac{2}{5}\]. Using the relationship between the variables, that is \[y=3x+1\] we can find the values of the variable \[y\] as follows,
For \[x=-1\], \[y=3(-1)+1=-2\].
For \[x=\dfrac{2}{5}\], \[y=3\left( \dfrac{2}{5} \right)+1=\dfrac{11}{5}\].
Hence the solution of the given simultaneous equation are \[x=-1,y=-2\] and \[x=\dfrac{2}{5},y=\dfrac{11}{5}\].
Note: Using the linear equation \[y=3x+1\], we get \[x=\dfrac{y-1}{3}\]. We can also substitute this relationship in the equation \[{{x}^{2}}+{{y}^{2}}=5\]. By this we will get an equation in terms of \[y\] by solving the equation, we will get the solution values of \[y\]. And then we can find the value of x for each \[y\].
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