Solved on Sep 25, 2023

Find $x$ and $y$ given $y=2x+3$ and $x^2-y+2k=0$ where $k\neq0$ . a) Show $x^2-2x+(2k-3)=0$ . b) Use the discriminant to find $k$ . c) Solve the equations for this $k$ .

STEP 1

Assumptions1. We have two simultaneous equations $y=x+3$ and $x^{}-y+k=0$ . . $k$ is a non-zero constant.
3. We need to show that $x^{}-x+(k-3)=0$ .
4. We need to use the discriminant to form an equation and solve for $k$ .
5. We need to solve the simultaneous equations for the value of $k$ obtained.

STEP 2

First, we substitute the value of $y$ from the first equation into the second equation.
$x^{2}-(2x+)+2k=0$

STEP 3

implify the equation by distributing the negative sign.
$x^{2}-2x-3+2k=0$

STEP 4

Rearrange the equation to match the form we need to prove.
$x^{2}-2x+(2k-3)=0$ This proves part (a) of the problem.

STEP 5

Given that $x^{2}-2x+(2k-3)$ has equal roots, we can use the discriminant of a quadratic equation, which is $b^{2}-4ac=0$ for equal roots.

STEP 6

In our equation $x^{2}-2x+(2k-3)=0$ , $a=1$ , $b=-2$ , and $c=2k-3$ .

STEP 7

Substitute these values into the discriminant equation.
$(-2)^{2}-4(1)(2k-3)=0$

STEP 8

implify the equation.
$4-8k+12=0$

STEP 9

Rearrange the equation to solve for $k$ .
$8k=16$

STEP 10

olve for $k$ .
$k=2$ This solves part (b) of the problem.

STEP 11

Now, we substitute $k=$ into the original simultaneous equations and solve for $x$ and $y$ .
The equations are $y=x+3$ $x^{}-y+4=0$

STEP 12

Substitute $y=2x+$ into the second equation.
$x^{2}-(2x+)+4=0$

STEP 13

implify the equation.
$x^{2}-2x+=0$

STEP 14

This is a perfect square trinomial, so we can factor it.
$(x-)^{2}=0$

STEP 15

olve for $x$ .
$x=$

STEP 16

Substitute $x=$ into the equation $y=2x+3$ to solve for $y$ .
$y=2()+3$

STEP 17

Calculate the value of $y$ .
$y=5$ So, for $k=2$ , the solutions to the simultaneous equations are $x=$ and $y=5$ .

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Find $x$ and $y$ given $y=2x+3$ and $x^2-y+2k=0$ where $k\neq0$ . a) Show $x^2-2x+(2k-3)=0$ . b) Use the discriminant to find $k$ . c) Solve the equations for this $k$ .

STEP 1

STEP 2

First, we substitute the value of $y$ from the first equation into the second equation.
$x^{2}-(2x+)+2k=0$

STEP 3

implify the equation by distributing the negative sign.
$x^{2}-2x-3+2k=0$

STEP 4

Rearrange the equation to match the form we need to prove.
$x^{2}-2x+(2k-3)=0$ This proves part (a) of the problem.

STEP 5

Given that $x^{2}-2x+(2k-3)$ has equal roots, we can use the discriminant of a quadratic equation, which is $b^{2}-4ac=0$ for equal roots.

STEP 6

In our equation $x^{2}-2x+(2k-3)=0$ , $a=1$ , $b=-2$ , and $c=2k-3$ .

STEP 7

Substitute these values into the discriminant equation.
$(-2)^{2}-4(1)(2k-3)=0$

STEP 8

implify the equation.
$4-8k+12=0$

STEP 9

Rearrange the equation to solve for $k$ .
$8k=16$

STEP 10

olve for $k$ .
$k=2$ This solves part (b) of the problem.

STEP 11

Now, we substitute $k=$ into the original simultaneous equations and solve for $x$ and $y$ .
The equations are $y=x+3$ $x^{}-y+4=0$

STEP 12

Substitute $y=2x+$ into the second equation.
$x^{2}-(2x+)+4=0$

STEP 13

implify the equation.
$x^{2}-2x+=0$

STEP 14

This is a perfect square trinomial, so we can factor it.
$(x-)^{2}=0$

STEP 15

olve for $x$ .
$x=$

STEP 16

Substitute $x=$ into the equation $y=2x+3$ to solve for $y$ .
$y=2()+3$

STEP 17

Calculate the value of $y$ .
$y=5$ So, for $k=2$ , the solutions to the simultaneous equations are $x=$ and $y=5$ .

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Find xxx and yyy given y=2x+3y=2x+3y=2x+3 and x2−y+2k=0x^2-y+2k=0x2−y+2k=0 where k≠0k\neq0k=0. a) Show x2−2x+(2k−3)=0x^2-2x+(2k-3)=0x2−2x+(2k−3)=0. b) Use the discriminant to find kkk. c) Solve the equations for this kkk.

STEP 1

STEP 2

First, we substitute the value of yyy from the first equation into the second equation.x2−(2x+)+2k=0x^{2}-(2x+)+2k=0x2−(2x+)+2k=0

STEP 3

implify the equation by distributing the negative sign.x2−2x−3+2k=0x^{2}-2x-3+2k=0x2−2x−3+2k=0

STEP 4

Rearrange the equation to match the form we need to prove.x2−2x+(2k−3)=0x^{2}-2x+(2k-3)=0x2−2x+(2k−3)=0This proves part (a) of the problem.

STEP 5

Given that x2−2x+(2k−3)x^{2}-2x+(2k-3)x2−2x+(2k−3) has equal roots, we can use the discriminant of a quadratic equation, which is b2−4ac=0b^{2}-4ac=0b2−4ac=0 for equal roots.

STEP 6

In our equation x2−2x+(2k−3)=0x^{2}-2x+(2k-3)=0x2−2x+(2k−3)=0, a=1a=1a=1, b=−2b=-2b=−2, and c=2k−3c=2k-3c=2k−3.

STEP 7

Substitute these values into the discriminant equation.(−2)2−4(1)(2k−3)=0(-2)^{2}-4(1)(2k-3)=0(−2)2−4(1)(2k−3)=0

STEP 8

implify the equation.4−8k+12=04-8k+12=04−8k+12=0

STEP 9

Rearrange the equation to solve for kkk.8k=168k=168k=16

STEP 10

olve for kkk.k=2k=2k=2This solves part (b) of the problem.

STEP 11

Now, we substitute k=k=k= into the original simultaneous equations and solve for xxx and yyy.The equations arey=x+3y=x+3y=x+3x−y+4=0x^{}-y+4=0x−y+4=0

STEP 12

Substitute y=2x+y=2x+y=2x+ into the second equation.x2−(2x+)+4=0x^{2}-(2x+)+4=0x2−(2x+)+4=0

STEP 13

implify the equation.x2−2x+=0x^{2}-2x+=0x2−2x+=0

STEP 14

This is a perfect square trinomial, so we can factor it.(x−)2=0(x-)^{2}=0(x−)2=0

STEP 15

olve for xxx.x=x=x=

STEP 16

Substitute x=x=x= into the equation y=2x+3y=2x+3y=2x+3 to solve for yyy.y=2()+3y=2()+3y=2()+3

STEP 17

Calculate the value of yyy.y=5y=5y=5So, for k=2k=2k=2, the solutions to the simultaneous equations are x=x=x= and y=5y=5y=5.

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Find $x$ and $y$ given $y=2x+3$ and $x^2-y+2k=0$ where $k\neq0$ . a) Show $x^2-2x+(2k-3)=0$ . b) Use the discriminant to find $k$ . c) Solve the equations for this $k$ .

First, we substitute the value of $y$ from the first equation into the second equation.
$x^{2}-(2x+)+2k=0$

implify the equation by distributing the negative sign.
$x^{2}-2x-3+2k=0$

Rearrange the equation to match the form we need to prove.
$x^{2}-2x+(2k-3)=0$ This proves part (a) of the problem.

Given that $x^{2}-2x+(2k-3)$ has equal roots, we can use the discriminant of a quadratic equation, which is $b^{2}-4ac=0$ for equal roots.

In our equation $x^{2}-2x+(2k-3)=0$ , $a=1$ , $b=-2$ , and $c=2k-3$ .

Substitute these values into the discriminant equation.
$(-2)^{2}-4(1)(2k-3)=0$

implify the equation.
$4-8k+12=0$

Rearrange the equation to solve for $k$ .
$8k=16$

olve for $k$ .
$k=2$ This solves part (b) of the problem.

Now, we substitute $k=$ into the original simultaneous equations and solve for $x$ and $y$ .
The equations are $y=x+3$ $x^{}-y+4=0$

Substitute $y=2x+$ into the second equation.
$x^{2}-(2x+)+4=0$

implify the equation.
$x^{2}-2x+=0$

This is a perfect square trinomial, so we can factor it.
$(x-)^{2}=0$

olve for $x$ .
$x=$

Substitute $x=$ into the equation $y=2x+3$ to solve for $y$ .
$y=2()+3$

Calculate the value of $y$ .
$y=5$ So, for $k=2$ , the solutions to the simultaneous equations are $x=$ and $y=5$ .