Subtracting the second equation from the first, we acquire \(\displaystyle 2x-5z=-11\).

Adding this equation to the third equation, we get \(\displaystyle 4x=8\).

Therefore, \(\displaystyle x=2\).

Solve for \(\displaystyle y\) by merging the equations so that \(\displaystyle x\) gets factored out.  To do so, multiply the lower equation by \(\displaystyle -9\) (so that \(\displaystyle 81x\) at the top is subtracted by \(\displaystyle -81x\))

\(\displaystyle (-9)9x + (-9)16y = (-9)25\)

\(\displaystyle (81x + 64y = 65) - (-81x - 144y = -225) =\)

\(\displaystyle 0x + (64 - 144)y = 65 - 225\)

\(\displaystyle -80y = -160\)

\(\displaystyle y =\frac{-160}{-80}=2\)

If the total milkshake profits were $246.05, and vanilla milkshakes cost $1.30 while chocolate and strawberry milkshakes cost $1.45, we can set up the equation 

\(\displaystyle 1.3a+1.45b=246.05\),

where a represents the number of vanilla milkshakes sold and b represents the number of chocolate and strawberry milkshakes sold. We also know that 179 total milkshakes were sold, so we can set up the equation 

\(\displaystyle a+b=179\).

If we solve for b, 

\(\displaystyle b=179-a\)

and substitute that value into our first equation 

\(\displaystyle 1.3a + 1.45(179-a)=246.05\)

we can solve for a and then b, and we get

\(\displaystyle a=90, b=89\)

This means that 90 vanilla milkshakes were sold and 89 chocolate and strawberry milkshakes were sold. If there were 27 more chocolate milkshakes sold than strawberry, we can set up the equation \(\displaystyle s+(s+27)=89\) where s represents the number of strawberry milkshakes sold, and s+27 represents the number of chocolate milkshakes sold. We can now solve for s. 

\(\displaystyle 2s+27=89\)

\(\displaystyle 2s=62\)

\(\displaystyle s=31\)

Therefore, 31 strawberry milkshakes were sold. 

In order to find the solution, you must set up a system of equations. We know that the product of two numbers, say x and y, is equal to 77.

Therefore, your first equation is

\(\displaystyle xy = 77\).

Next, we know that one number is 3 less than twice the value of the other number, or that

\(\displaystyle x = 2y - 3\).

Once we have our two equations, we can use substitution in order to eliminate a variable and solve for the remaining variable.

We know that,

\(\displaystyle x =2y - 3\),

so instead of

\(\displaystyle xy = 77\) 

we can say that (2y - 3) times y equals 77, or

\(\displaystyle y(2y - 3) = 77\).

To solve for y we need to factor.

\(\displaystyle 2y^2 - 3y - 77 = 0\) 

becomes

\(\displaystyle (2y +11)(y - 7) = 0\).

Because y must be positive, we know that y must equal 7. Substituting 7 for y into our original equation (xy = 77), we learn that x equals 11.

Therefore \(\displaystyle x = 11\) and \(\displaystyle y = 7\) and the sum of our two numbers equals 18.

It is necessary to find the vertices of the triangle,each of which is the intersection of two lines.

The lines of the equations \(\displaystyle x = 1\) and \(\displaystyle y = -1\) can be immediately seen to intersect at \(\displaystyle (1, -1)\).

The intersection of the lines of equations \(\displaystyle x = 1\) and \(\displaystyle y =3x + 7\) can be found by substituting 1 for \(\displaystyle x\) in the second equation and evaluating \(\displaystyle y\):

\(\displaystyle y =3 \cdot 1 + 7 = 3 + 7 = 10\)

Their point of intersection is at \(\displaystyle (1, 10)\).

Similarly, the intersection of the lines of equations \(\displaystyle y = -1\) and \(\displaystyle y =3x + 7\) can be found by substituting \(\displaystyle -1\) for \(\displaystyle y\) in the second equation and solving for \(\displaystyle x\):

\(\displaystyle 3x + 7 = -1\)

\(\displaystyle 3x + 7 - 7 = -1 - 7\)

\(\displaystyle 3x = -8\)

\(\displaystyle \frac{3x}{3} = \frac{-8}{3}\)

\(\displaystyle x=- \frac{8}{3} = -2\frac{2}{3}\)

Their point of intersection is at \(\displaystyle \left ( -2 \frac{2}{3}, -1 \right )\).

The lines in question are graphed below, and the triangle they bound is shaded:

The triangle is right, so its area is half the product of the lengths of its legs, which here are its vertical and horizontal sides. The length of its horizontal leg is the difference of the \(\displaystyle x\)-coordinates of its endpoints:

\(\displaystyle 1 - \left ( -2 \frac{2}{3}\right ) = 3 \frac{2}{3}\)

The length of its vertical leg is the difference of the \(\displaystyle y\)-coordinates of its endpoints:

\(\displaystyle 10 - (-1) = 11\)

Half the product of these is the area:

\(\displaystyle A =\frac{1}{2} ab = \frac{1}{2} \cdot 3 \frac{2}{3} \cdot 11 = 20 \frac{1}{6}\)

Substitution

Solve the second equation for x:

\(\displaystyle x = 4y - 23\)

Substitute this expression for x in the first equation:

\(\displaystyle 3(4y - 23) - 2y = 1\)

Solve for y:

\(\displaystyle \\12y - 69 - 2y = 1 \\10y = 70 \\ y = 7\)

Substitute this value for y in any equation and solve for x.

\(\displaystyle \\x = 4(7) - 23 = 28 - 23 = 5 \\x + y = 5 + 7 = 12\)