Math / Algebra

Question9. In the following set of nonhomogeneous equations if $z=1$ , then the value of $a$ is: $\begin{array}{c} x+3 y+a z=2 \\ y+z=-1 \\ 5 x-2 y-z=3 \end{array}$ a. 25 b. 3 c. -3 d. -8 e. 8

Studdy Solution

STEP 1

What is this asking? We're given a system of equations and told $z=1$ , and we need to find the value of $a$ . Watch out! Don't mix up your variables!
Keep track of your $x$ 's, $y$ 's, and $z$ 's.
Also, remember we're solving for $a$ , not $x$ or $y$ !

STEP 2

1. Solve for $y$ .
2. Solve for $x$ .
3. Solve for $a$ .

STEP 3

Alright, let's kick things off with the second equation because it looks the simplest: $y + z = -1$ .
We know that $z = 1$ , so let's plug that in: $y + 1 = -1$ .

STEP 4

Now, we want to isolate $y$ .
To do this, we can add $-1$ to both sides of the equation.
This gives us $y + 1 + (-1) = -1 + (-1)$ , which simplifies to $y = \mathbf{-2}$ .
Awesome! We found $y$ !

STEP 5

Next up, let's tackle the third equation: $5x - 2y - z = 3$ .
We already know $z = 1$ and $y = -2$ , so let's substitute those values in: $5x - 2(-2) - 1 = 3$ .

STEP 6

Let's simplify things a bit: $5x + 4 - 1 = 3$ , which becomes $5x + 3 = 3$ .

STEP 7

Now, we want to isolate $x$ .
Let's add $-3$ to both sides: $5x + 3 + (-3) = 3 + (-3)$ , which simplifies to $5x = 0$ .

STEP 8

Finally, to completely isolate $x$ , we multiply both sides by $\frac{1}{5}$ : $\frac{1}{5} \cdot 5x = \frac{1}{5} \cdot 0$ , which gives us $x = \mathbf{0}$ .
Boom! We've got $x$ now!

STEP 9

Time for the grand finale!
We're going to use the first equation: $x + 3y + az = 2$ .
We know $x = 0$ , $y = -2$ , and $z = 1$ , so let's plug those values in: $0 + 3(-2) + a(1) = 2$ .

STEP 10

Let's simplify: $-6 + a = 2$ .

STEP 11

To solve for $a$ , we add $6$ to both sides of the equation: $-6 + a + 6 = 2 + 6$ , which simplifies to $a = \mathbf{8}$ .
We did it!

STEP 12

The value of $a$ is 8, so the answer is (e).

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Question9. In the following set of nonhomogeneous equations if $z=1$ , then the value of $a$ is: $\begin{array}{c} x+3 y+a z=2 \\ y+z=-1 \\ 5 x-2 y-z=3 \end{array}$ a. 25 b. 3 c. -3 d. -8 e. 8

Studdy Solution

STEP 1

What is this asking? We're given a system of equations and told $z=1$ , and we need to find the value of $a$ . Watch out! Don't mix up your variables!
Keep track of your $x$ 's, $y$ 's, and $z$ 's.
Also, remember we're solving for $a$ , not $x$ or $y$ !

STEP 2

1. Solve for $y$ .
2. Solve for $x$ .
3. Solve for $a$ .

STEP 3

Alright, let's kick things off with the second equation because it looks the simplest: $y + z = -1$ .
We know that $z = 1$ , so let's plug that in: $y + 1 = -1$ .

STEP 4

Now, we want to isolate $y$ .
To do this, we can add $-1$ to both sides of the equation.
This gives us $y + 1 + (-1) = -1 + (-1)$ , which simplifies to $y = \mathbf{-2}$ .
Awesome! We found $y$ !

STEP 5

Next up, let's tackle the third equation: $5x - 2y - z = 3$ .
We already know $z = 1$ and $y = -2$ , so let's substitute those values in: $5x - 2(-2) - 1 = 3$ .

STEP 6

Let's simplify things a bit: $5x + 4 - 1 = 3$ , which becomes $5x + 3 = 3$ .

STEP 7

Now, we want to isolate $x$ .
Let's add $-3$ to both sides: $5x + 3 + (-3) = 3 + (-3)$ , which simplifies to $5x = 0$ .

STEP 8

Finally, to completely isolate $x$ , we multiply both sides by $\frac{1}{5}$ : $\frac{1}{5} \cdot 5x = \frac{1}{5} \cdot 0$ , which gives us $x = \mathbf{0}$ .
Boom! We've got $x$ now!

STEP 9

Time for the grand finale!
We're going to use the first equation: $x + 3y + az = 2$ .
We know $x = 0$ , $y = -2$ , and $z = 1$ , so let's plug those values in: $0 + 3(-2) + a(1) = 2$ .

STEP 10

Let's simplify: $-6 + a = 2$ .

STEP 11

To solve for $a$ , we add $6$ to both sides of the equation: $-6 + a + 6 = 2 + 6$ , which simplifies to $a = \mathbf{8}$ .
We did it!

STEP 12

The value of $a$ is 8, so the answer is (e).

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Question9. In the following set of nonhomogeneous equations if z=1z=1z=1, then the value of aaa is: x+3y+az=2y+z=−15x−2y−z=3\begin{array}{c} x+3 y+a z=2 \\ y+z=-1 \\ 5 x-2 y-z=3 \end{array}x+3y+az=2y+z=−15x−2y−z=3​ a. 25 b. 3 c. -3 d. -8 e. 8

Studdy Solution

STEP 1

What is this asking? We're given a system of equations and told z=1z=1z=1, and we need to find the value of aaa. Watch out! Don't mix up your variables!Keep track of your xxx's, yyy's, and zzz's.Also, remember we're solving for aaa, not xxx or yyy!

STEP 2

1. Solve for yyy.2. Solve for xxx.3. Solve for aaa.

STEP 3

Alright, let's **kick things off** with the second equation because it looks the simplest: y+z=−1y + z = -1y+z=−1.We know that z=1z = 1z=1, so let's **plug that in**: y+1=−1y + 1 = -1y+1=−1.

STEP 4

Now, we want to **isolate** yyy.To do this, we can **add** −1-1−1 to both sides of the equation.This gives us y+1+(−1)=−1+(−1)y + 1 + (-1) = -1 + (-1)y+1+(−1)=−1+(−1), which simplifies to y=−2y = \mathbf{-2}y=−2.Awesome! We found yyy!

STEP 5

Next up, let's **tackle** the third equation: 5x−2y−z=35x - 2y - z = 35x−2y−z=3.We already know z=1z = 1z=1 and y=−2y = -2y=−2, so let's **substitute** those values in: 5x−2(−2)−1=35x - 2(-2) - 1 = 35x−2(−2)−1=3.

STEP 6

Let's **simplify** things a bit: 5x+4−1=35x + 4 - 1 = 35x+4−1=3, which becomes 5x+3=35x + 3 = 35x+3=3.

STEP 7

Now, we want to **isolate** xxx.Let's **add** −3-3−3 to both sides: 5x+3+(−3)=3+(−3)5x + 3 + (-3) = 3 + (-3)5x+3+(−3)=3+(−3), which simplifies to 5x=05x = 05x=0.

STEP 8

Finally, to **completely isolate** xxx, we **multiply** both sides by 15\frac{1}{5}51​: 15⋅5x=15⋅0\frac{1}{5} \cdot 5x = \frac{1}{5} \cdot 051​⋅5x=51​⋅0, which gives us x=0x = \mathbf{0}x=0.Boom! We've got xxx now!

STEP 9

Time for the **grand finale**!We're going to use the first equation: x+3y+az=2x + 3y + az = 2x+3y+az=2.We know x=0x = 0x=0, y=−2y = -2y=−2, and z=1z = 1z=1, so let's **plug** those values in: 0+3(−2)+a(1)=20 + 3(-2) + a(1) = 20+3(−2)+a(1)=2.

STEP 10

Let's **simplify**: −6+a=2-6 + a = 2−6+a=2.

STEP 11

To **solve for** aaa, we **add** 666 to both sides of the equation: −6+a+6=2+6-6 + a + 6 = 2 + 6−6+a+6=2+6, which simplifies to a=8a = \mathbf{8}a=8.We did it!

STEP 12

The value of aaa is 8, so the answer is (e).

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Question9. In the following set of nonhomogeneous equations if $z=1$ , then the value of $a$ is: $\begin{array}{c} x+3 y+a z=2 \\ y+z=-1 \\ 5 x-2 y-z=3 \end{array}$ a. 25 b. 3 c. -3 d. -8 e. 8

What is this asking? We're given a system of equations and told $z=1$ , and we need to find the value of $a$ . Watch out! Don't mix up your variables!
Keep track of your $x$ 's, $y$ 's, and $z$ 's.
Also, remember we're solving for $a$ , not $x$ or $y$ !

1. Solve for $y$ .
2. Solve for $x$ .
3. Solve for $a$ .

Alright, let's kick things off with the second equation because it looks the simplest: $y + z = -1$ .
We know that $z = 1$ , so let's plug that in: $y + 1 = -1$ .

Now, we want to isolate $y$ .
To do this, we can add $-1$ to both sides of the equation.
This gives us $y + 1 + (-1) = -1 + (-1)$ , which simplifies to $y = \mathbf{-2}$ .
Awesome! We found $y$ !

Next up, let's tackle the third equation: $5x - 2y - z = 3$ .
We already know $z = 1$ and $y = -2$ , so let's substitute those values in: $5x - 2(-2) - 1 = 3$ .

Let's simplify things a bit: $5x + 4 - 1 = 3$ , which becomes $5x + 3 = 3$ .

Now, we want to isolate $x$ .
Let's add $-3$ to both sides: $5x + 3 + (-3) = 3 + (-3)$ , which simplifies to $5x = 0$ .

Finally, to completely isolate $x$ , we multiply both sides by $\frac{1}{5}$ : $\frac{1}{5} \cdot 5x = \frac{1}{5} \cdot 0$ , which gives us $x = \mathbf{0}$ .
Boom! We've got $x$ now!

Time for the grand finale!
We're going to use the first equation: $x + 3y + az = 2$ .
We know $x = 0$ , $y = -2$ , and $z = 1$ , so let's plug those values in: $0 + 3(-2) + a(1) = 2$ .

Let's simplify: $-6 + a = 2$ .

To solve for $a$ , we add $6$ to both sides of the equation: $-6 + a + 6 = 2 + 6$ , which simplifies to $a = \mathbf{8}$ .
We did it!

The value of $a$ is 8, so the answer is (e).