Question

How‌ ‌do‌ ‌you‌ ‌solve‌ ‌the‌ ‌system‌ ‌$3{{x}^{2}}-5{{y}^{2}}+2y=45$‌ ‌and‌ ‌$y=2x+10$?‌

Answer 1

In this problem, we have to solve the given system of equations to find the value of x and y. We can substitute one of the equations in the other. We will get a quadratic equation on simplifying it. We can solve the quadratic equation using the quadratic formula to solve for x and we can substitute the value of x in one of the solutions to get the value of y.

Complete step by step solution:
We know that the given system of equations to be solved,
$3{{x}^{2}}-5{{y}^{2}}+2y=45$…… (1)
$y=2x+10$…… (2)
We can now substitute the equation (2) in the equation (1), we get
$\Rightarrow 3{{x}^{2}}-5{{\left( 2x+10 \right)}^{2}}+2\left( 2x+10 \right)=45$
Now we can simplify the above step using algebraic square formula, we get
$\Rightarrow 3{{x}^{2}}-5\left( 4{{x}^{2}}+40x+100 \right)+4x+20-45=0$
We can further simplify the above step we get
$\Rightarrow 3{{x}^{2}}-20{{x}^{2}}-200x-500+4x+20-45=0$
$\Rightarrow -17{{x}^{2}}-196x-525=0$
We can multiply -1 on both sides, we get
$\Rightarrow 17{{x}^{2}}+196x+525=0$ …….. (3)
Now we have a quadratic equation. We can solve this quadratic equation to find the value of x by using the quadratic formula.
We also know that a quadratic equation in standard form is,
$a{{x}^{2}}+bx+c=0$ ……. (4)
We can now compare the two equations (3) and (4), we get
a = 17, b = 196, c = 525.
We know that the quadratic formula for the standard form $a{{x}^{2}}+bx+c=0$ is
$x=\dfrac{-b\pm \sqrt{{{\left( b \right)}^{2}}-4\times a\times \left( c \right)}}{2a}$
Now we can substitute the value of a, b, c in the above formula, we get
$\Rightarrow x=\dfrac{-196\pm \sqrt{{{\left( 196 \right)}^{2}}-4\times 17\times \left( 525 \right)}}{2\times 17}$
Now we can simplify the above step, we get

& \Rightarrow x=\dfrac{-196\pm \sqrt{38416-35700}}{34} \\\ & \Rightarrow x=\dfrac{-196\pm 52.21}{34} \\\ \end{aligned}$$ Now we can simplify the above step using calculator, we get $$\Rightarrow x=-4.2,-7.3$$ Now we can substitute the above values in equation (2), we get $$\begin{aligned} & \Rightarrow y=2\left( -4.2 \right)+10=1.6 \\\ & \Rightarrow y=2\left( -7.3 \right)+10=-4.6 \\\ \end{aligned}$$ **Therefore, the value of $$\left( x,y \right)$$ is $$\left( -4.2,1.6 \right)$$ and $$\left( -7.3,-4.6 \right)$$.** **Note:** Students make mistakes in the quadratic formula part, which we should concentrate on. We may feel difficult to solve these types of complicated root values in the quadratic formula for which we can use the scientific calculators to get the exact values.