Math / Algebra

Question $3215 / 25509156 /$ A/694AdB8dS6abeB3a9c1 1/2b401747 quesmon Find all solutions of the system of equations algebraically. Write your solutio points. $\begin{array}{c} y=x^{2}+3 x-41 \\ 16=3 x-y \end{array}$
Answer Attempt 2 out of 2
Two solutions $(5 i, 15 i-16)$ and $(-5 i-15 I-16)$ $\square$ Submit Answer

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ and $y$ that make both equations true at the same time! Watch out! Don't forget that when we substitute, we need to be super careful with our signs and make sure we solve for both $x$ and $y$ !

STEP 2

1. Prep the second equation
2. Substitute and solve for $x$
3. Find the matching $y$ values

STEP 3

Let's get $y$ by itself in the second equation, $16 = 3x - y$ , so it's ready for substitution.
We can add $y$ to both sides to get $16 + y = 3x$ .
Then, subtract 16 from both sides to get $y = 3x - 16$ .
Perfect!

STEP 4

Now, let's substitute this expression for $y$ into the first equation: $y = x^2 + 3x - 41$ becomes $(3x - 16) = x^2 + 3x - 41$ .

STEP 5

We want to solve for $x$ , so let's get everything on one side of the equation.
Subtract $3x$ from both sides: $-16 = x^2 - 41$ .
Now, add 41 to both sides to get $x^2 = 25$ .
Taking the square root of both sides gives us $x = 5$ and $x = -5$ .
We have two solutions for $x$ !

STEP 6

Let's plug $x = 5$ back into the equation $y = 3x - 16$ .
So, $y = 3 \cdot 5 - 16$ , which simplifies to $y = 15 - 16 = -1$ .
Our first solution point is $(5, -1)$ .

STEP 7

Now, let's plug $x = -5$ into $y = 3x - 16$ .
This gives us $y = 3 \cdot (-5) - 16$ , which simplifies to $y = -15 - 16 = -31$ .
Our second solution point is $(-5, -31)$ .

STEP 8

The solutions to the system of equations are $(5, -1)$ and $(-5, -31)$ !

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Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ and $y$ that make both equations true at the same time! Watch out! Don't forget that when we substitute, we need to be super careful with our signs and make sure we solve for both $x$ and $y$ !

STEP 2

1. Prep the second equation
2. Substitute and solve for $x$
3. Find the matching $y$ values

STEP 3

Let's get $y$ by itself in the second equation, $16 = 3x - y$ , so it's ready for substitution.
We can add $y$ to both sides to get $16 + y = 3x$ .
Then, subtract 16 from both sides to get $y = 3x - 16$ .
Perfect!

STEP 4

Now, let's substitute this expression for $y$ into the first equation: $y = x^2 + 3x - 41$ becomes $(3x - 16) = x^2 + 3x - 41$ .

STEP 5

We want to solve for $x$ , so let's get everything on one side of the equation.
Subtract $3x$ from both sides: $-16 = x^2 - 41$ .
Now, add 41 to both sides to get $x^2 = 25$ .
Taking the square root of both sides gives us $x = 5$ and $x = -5$ .
We have two solutions for $x$ !

STEP 6

Let's plug $x = 5$ back into the equation $y = 3x - 16$ .
So, $y = 3 \cdot 5 - 16$ , which simplifies to $y = 15 - 16 = -1$ .
Our first solution point is $(5, -1)$ .

STEP 7

Now, let's plug $x = -5$ into $y = 3x - 16$ .
This gives us $y = 3 \cdot (-5) - 16$ , which simplifies to $y = -15 - 16 = -31$ .
Our second solution point is $(-5, -31)$ .

STEP 8

The solutions to the system of equations are $(5, -1)$ and $(-5, -31)$ !

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Studdy Solution

STEP 1

What is this asking? We need to find the values of xxx and yyy that make both equations true at the same time! Watch out! Don't forget that when we substitute, we need to be super careful with our signs and make sure we solve for *both* xxx and yyy!

STEP 2

1. Prep the second equation2. Substitute and solve for xxx3. Find the matching yyy values

STEP 3

Let's get yyy by itself in the second equation, 16=3x−y16 = 3x - y16=3x−y, so it's ready for substitution.We can add yyy to both sides to get 16+y=3x16 + y = 3x16+y=3x.Then, subtract 16 from both sides to get y=3x−16y = 3x - 16y=3x−16.Perfect!

STEP 4

Now, let's **substitute** this expression for yyy into the first equation: y=x2+3x−41y = x^2 + 3x - 41y=x2+3x−41 becomes (3x−16)=x2+3x−41(3x - 16) = x^2 + 3x - 41(3x−16)=x2+3x−41.

STEP 5

STEP 6

Let's plug x=5x = 5x=5 back into the equation y=3x−16y = 3x - 16y=3x−16.So, y=3⋅5−16y = 3 \cdot 5 - 16y=3⋅5−16, which simplifies to y=15−16=−1y = 15 - 16 = -1y=15−16=−1.Our first solution point is (5,−1)(5, -1)(5,−1).

STEP 7

Now, let's plug x=−5x = -5x=−5 into y=3x−16y = 3x - 16y=3x−16.This gives us y=3⋅(−5)−16y = 3 \cdot (-5) - 16y=3⋅(−5)−16, which simplifies to y=−15−16=−31y = -15 - 16 = -31y=−15−16=−31.Our second solution point is (−5,−31)(-5, -31)(−5,−31).

STEP 8

The solutions to the system of equations are (5,−1)(5, -1)(5,−1) and (−5,−31)(-5, -31)(−5,−31)!

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to find the values of $x$ and $y$ that make both equations true at the same time! Watch out! Don't forget that when we substitute, we need to be super careful with our signs and make sure we solve for both $x$ and $y$ !

1. Prep the second equation
2. Substitute and solve for $x$
3. Find the matching $y$ values

Let's get $y$ by itself in the second equation, $16 = 3x - y$ , so it's ready for substitution.
We can add $y$ to both sides to get $16 + y = 3x$ .
Then, subtract 16 from both sides to get $y = 3x - 16$ .
Perfect!

Now, let's substitute this expression for $y$ into the first equation: $y = x^2 + 3x - 41$ becomes $(3x - 16) = x^2 + 3x - 41$ .

Let's plug $x = 5$ back into the equation $y = 3x - 16$ .
So, $y = 3 \cdot 5 - 16$ , which simplifies to $y = 15 - 16 = -1$ .
Our first solution point is $(5, -1)$ .

Now, let's plug $x = -5$ into $y = 3x - 16$ .
This gives us $y = 3 \cdot (-5) - 16$ , which simplifies to $y = -15 - 16 = -31$ .
Our second solution point is $(-5, -31)$ .

The solutions to the system of equations are $(5, -1)$ and $(-5, -31)$ !