Answer:
x = 8
y = -7
Step-by-step explanation:
This is a system of equations called simultaneous equations. We shall solve it by elimination method Step 1We shall label the equations (1) and (2)−3y−4x=−11.....(1)3y−5x=−61......(2)Step 2Multiply each term in equation (1) by 1 to give equation (3)1(-3y-4x=-11).....(1)-3y-4x=-11....(3)Step 3Multiply each term in equation 2 by -1 to give equation (4)-1(3y−5x=−61)......(2)-3y+5x=61.....(4)Step 4-3y-4x=-11....(3)-3y+5x=61.....(4)Subtract each term in equation (3) from each term in equation (4)-3y-(-3y)+5x-(-4x)=61-(-11)-3y+3y+5x+4x=61+110+9x=729x=72Step 5Divide both sides of the equation by 9, the coefficient of the unknown variable x to find the value of x 9x/9 = 72/9x = 8Step 6Put in x = 8 into equation (2)3y−5x=−61......(2)3y-5(8)=-613y-40=-61Collect like terms by adding 40 to both sides of the equation 3y-40+40=-61+403y=-21Divide both sides by 3, the coefficient of y to find the value of y 3y/3=-21/3y=-7Therefore, the values of x and y that satisfy the equations are 8 and -7 respectively
Solutions
In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.<span>
Calculations
</span>⇒ <span>Rewrite the linear equations above as a matrix
</span>
⇒ Apply to Row₂ : Row₂ - 2 <span>Row₁
</span>
⇒ <span>Simplify rows
</span>
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂
⇒ <span>Simplify rows
</span>
⇒ <span>Therefore,
</span>

<span>
</span>
Answer:
8070
Step-by-step explanation:
Given:
8.07
x
10
3
8.07
×
10
3
=
8.07
×
1000
=
8070
D because if you multiply y 8 times it will no longer be doubling the area
Greetings.
The range is the set of y-value.
The range starts from the minimum point to maximum point.
Our minimum point starts at 0 and maximum point starts less than infinity.
Therefore the range is 0<=y<+inf
However, we do not often write that, although it is right.
Therefore we write as y≥0
Thus, the answer is B choice.