Question

2x+y=9 ,3x²+2y²+x-y=45
solve simultaneously ​

Answer 1

Hi,

Explanation:

To solve the system of equations simultaneously, you can use the method of substitution or elimination.

Let's use the elimination method:

Given system:

`\left\{\begin{array}{ccc}2x+y&=&9\\3x^2+2y^2+x-y&=&45\\\end {array} \right.\\`

First, isolate y in the first equation .

And substitute that expression for y into the second equation.

`\left\{\begin{array}{ccc}y&=&-2x+9\\3x^2+2(-2x+9)^2+x-(-2x+9)&=&45\\\end {array} \right.\\`

Simplify the equation, expand and collect like terms:

`\left\{\begin{array}{ccc}y&=&-2x+9\\3x^2+2(4x^2-36x+81)+x+2x-9)&=&45\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&-2x+9\\11x^2-69x-108&=&0\\\end {array} \right.\\\\`

Now, you can solve this quadratic equation for x.

You can either factor it or use the quadratic formula.

`11x^2-33x-36x+108=0\\11x(x-3)-36(x-3)=0\\(x-3)(11x-36)=0\\`

This gives two possible solutions for x:

`x=3\ or\ x=(36)/(11) \\`

Now that you have the values of x, you can find the corresponding values of y using the expression y=9−2x.

Plug each x value into this expression to find the corresponding y values.

`y=9-2*3=3\ or \ y=9-2*(36)/(11) =(27)/(11) \\\\`

So you'll have two pairs of solutions :

`\left\{\begin{array}{ccc}x&=&3\\y&=&3\\\end {array} \right.\ or\ \left\{\begin{array}{ccc}x&=&(36)/(11)\\\\y&=&(27)/(11)\\\end {array} \right.`

Answer 2

Step-by-step explanation:Let’s start with the first:x = 3+2yx^2 + 2y^2 = 27Firstly, we need to align both equations so that they use similar terms, we can create an equation that contains x^2 and y^2 by squaring both sides of x = 3+ 2y(x)^2 = (3+ 2y)^2x^2 = 4y^2 + 12y + 9Then, we can get both equations to parallel each other by altering x^2 + 2y^2 = 27 so that x^2 is alone:x^2 + 2y^2 = 27 therefore x^2 = -2y^2 +27No we can subtract one equation from the other:x^2 = 4y^2 + 12y + 9x^2 = -2y^2 + 27Therefore0 = 6y^2 + 12y - 18This equation can be simplified by dividing all the elements by 6, leaving:y^2 + 2y - 3 = 0This factorises to:(y + 3)(y - 1) = 0Which results in two y values: -3 and 1From these values, you can create two values of x by using the equation x = 3 + 2y:Let y = -3:x = 3 + 2(-3)x = 3 - 6 = -3This can be verified using the other original equation:x^2 + 2y^2 = 27(-3)^2 + 2(-3)^2 = 279 + 2(9) = 279 + 18 = 2727 = 27For the other value:Let y = 1:x = 3 + 2yx = 3 + 2(1)x = 3 + 2 = 5Verify:x^2 + 2y^2 = 27(5)^2 + 2(1)^2 = 2725 + 2 = 2727 = 27