Question

How do you solve the system \[{{x}^{2}}+{{y}^{2}}=25,x+2y=10\]?

Answer 1

Hint: We will solve this using substitution method. First we will solve the linear equation for any one variable. Then we will substitute that variable in the quadratic equation and then solve that for roots. Then using those values we can find the other variable. We will get two values for each variable x and y.

Complete step by step solution:
Given equations are
\[{{x}^{2}}+{{y}^{2}}=25\]
\[x+2y=10\]
First we have to solve the linear equation for any one variable.
Our linear equation is
\[x+2y=10\]
Here we will solve for variable x.
So we have to group the terms in such a way that only x contains terms on LHS side and remaining on RHS side.
To rearrange the terms we have to subtract \[2y\] on both sides of the equation.
\[\Rightarrow x+2y-2y=10-2y\]
By simplifying we will get
\[\Rightarrow x=10-2y\]
Now we substitute this x value in the quadratic equation.
By substituting we will get
\[\Rightarrow {{\left( 10-2y \right)}^{2}}+{{y}^{2}}=25\]
Now we have to simplify the equation.
We will get
\[\Rightarrow 100+4{{y}^{2}}-40y+{{y}^{2}}=25\]
By rearranging the terms we will get
\[\Rightarrow 5{{y}^{2}}-40y+100=25\]
Now we can see that all the terms are factors of 5. So we divide the equation with 5 on both sides.
 \[\Rightarrow \dfrac{5{{y}^{2}}-40y+100}{5}=\dfrac{25}{5}\]
We will get
\[\Rightarrow {{y}^{2}}-8y+20=5\]
Now we subtract 5 from both sides of the equation.
We will get
\[\Rightarrow {{y}^{2}}-8y+15=0\]
So we got this quadratic equation. By solving this we can find roots for y.
Now we will solve the quadratic equation by factoring we will get
\[\Rightarrow {{y}^{2}}-3y-5y+15=0\]
By taking common terms we will get
\[\Rightarrow y\left( y-3 \right)-5\left( y-3 \right)\]
So the factors are
\[\Rightarrow \left( y-3 \right)\left( y-5 \right)\]
Then the roots are
\[\Rightarrow y=3,y=5\]
Now we will calculate x value for each y value.
For \[y=3\] the x value we get by substituting in first equation is
\[\Rightarrow x+2\left( 3 \right)=10\]
By simplifying we get
\[\Rightarrow x+6=10\]
Now subtracting 6 from both sides we will get
\[\Rightarrow x=4\]
Now we will calculate for \[y=5\] we will get
\[\Rightarrow x+2\left( 5 \right)=10\]
By simplifying
\[\Rightarrow x+10=10\]
\[\Rightarrow x=0\].

So the solutions for the given system are \[\left( 0,5 \right)\left( 4,3 \right)\].

Note: We can also start by solving with y but here we have chosen x because it’s coefficient is 1 so that it is easy to solve. Also we can use quadratic formulas to solve the quadratic equation. The quadratic equation will form a circle and the line will intersect at the roots we got.