Question

7.

Answer 1

Answer:

7.The solution to the set of equation in the form (-5,-3).

8.Multiply equation A by -4 used to eliminate  the z- term.

9.Step 2: $-4y=-16+8z$ { equation A in step1 is simplified}.

10. $p= d+2$

$p=d-1$.

Step-by-step explanation:

7. Two equation are given below:

$a-3b=4$

$a=b-2$

II eqaution can be write as

$a-b=-2$

Subtracting equation II from  equation I then we get

$-2b=6$

By division property of equality

$b=\frac{6}{-2}$

By simplification we get

$b=-3$

Substitute the value of b in equation I then we get

$a-3(-3)=4$

$a+9=4$

$a=4-9$

$a=-5$

Hence, the solution of the set of equation is (-5,-3).

8. Equation A: $x+z=6$

Equation B: $2x+4z=1$

Equation A is multiplied by -4 then we get

Equation A:$-4x-4z=-24$

Adding both equation A and B then we get

$-2x=-23$

Answer: Multiply equation A by -4 to eliminate the z-term.

9.Equation A: $y=4-2z$

Equation B:$4y=2-4z$

Step1 :$-4(y)=-4(4-2z)$

Equation A is multiplied by -4

$4y=2-4z$ [equation B]

Step 2: $-4y=-16+8z$

$4y=2-4z$ [equation B]

Equation A in step1 s simplified .

Step3: $0= -14+4z$

Equations in step 2 are added.

Step 4: $4z=14$

Step5: $z=\frac{7}{2}$

Hence, in step 2 student did make first an error.

10. Given

Variable p is more than variable d

We can write in algebraic expression

$p=d+2$

Variable p is also 1 less than variable d.

Then the algebraic expression

$p=d-1$

Hence, $p=d+2$

$p=d-1$

Answer 2

7. b - 2 - 3b = 4
   -2b = 6
    b = -3
    a = -5
solution is (-5,-3) <==
8. multiply A by -4...this eliminates the z's when added <==
9. first error...step 2....he didn't distribute correctly <==
10. p = d + 2 : p = d - 1 <==