Let's first look at the 2nd equation. All three terms in \(\displaystyle \dpi{100} \small 7x-14y=21\) can be divided by 7.  Then \(\displaystyle \dpi{100} \small x-2y=3\)  We can isolate x to get \(\displaystyle \dpi{100} \small x=3+2y\)

Now let's plug \(\displaystyle \dpi{100} \small x=3+2y\) into the 1st equation, \(\displaystyle \dpi{100} \small 4x+3y=7:\)

\(\displaystyle \dpi{100} \small 4\left ( 3+2y \right )+3y=7\)

\(\displaystyle \dpi{100} \small 12+8y+3y=7\)

\(\displaystyle \dpi{100} \small 11y=-5\)

\(\displaystyle \dpi{100} \small y=\frac{-5}{11}\)  Now let's plug our y-value into \(\displaystyle \dpi{100} \small x=3+2y\) to solve for y:

\(\displaystyle \dpi{100} \small x=3+2\left\left ( \frac{-5}{11} \right )=3-\frac{10}{11}=\frac{33}{11}-\frac{10}{11}=\frac{23}{11}\)

So \(\displaystyle \dpi{100} \small x=\frac{23}{11}\and\ y=\frac{-5}{11}\)

\(\displaystyle 4x+3y=19\)

\(\displaystyle 5x+4y = 23\)

Subtract the first equation from the second:

\(\displaystyle x+y = 4\Rightarrow y=4-x\)

Now we can substitute this into either equation. We'll plug it into the first equation here:

\(\displaystyle 4x + 3(4-x) = 19\Rightarrow 4x+12-3x = 19\Rightarrow x+12 = 19\)

Thus we get \(\displaystyle x=7\) and \(\displaystyle y=4-(7) = -3\).

Therefore \(\displaystyle x\) is positive and \(\displaystyle y\) is negative.

For this problem we can use the elimination method to solve for one of our variables. We do this my multiplying our first equation by -2.

\(\displaystyle -2(2x+y=7)=-4x-2y=-14\)

From here we can combine this equation with our second equation given in the question and solve for x.

\(\displaystyle -4x-2y=-14\)

\(\displaystyle +(3x+2y=13)\)

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\(\displaystyle -x=-1\)

\(\displaystyle x=1\)

Now we plug 1 back into our original equation and solve for y.

\(\displaystyle 3(1)+2y=13\)

\(\displaystyle 2y=10\)

\(\displaystyle y=5\)

Therefore,

\(\displaystyle x+y=1+5=6\)

The correct answer is \(\displaystyle \left(2,-\frac{1}{6}\right)\)

There are a few ways of solving this. The method I will use is the method of elimination.

\(\displaystyle 2x+6y=3\) (Start)

\(\displaystyle x+6y=1\)

\(\displaystyle x+0y=2\)

\(\displaystyle x+6y=1\)(Multiply the 2nd equation by -1 and add the result to the first equation, combining like terms. Now the top equation simplifies to \(\displaystyle x=2\)

Now that we have one of the variables solved for, we can plug \(\displaystyle x=2\) into either of the original equations, and we can get our \(\displaystyle y\), Let's use the 2nd equation.

\(\displaystyle (2)+6y=1\)

\(\displaystyle 6y=-1\)

\(\displaystyle y=-\frac{1}{6}\)

Hence the point of intersection of the two lines is \(\displaystyle \left(2,-\frac{1}{6}\right)\).

Let \(\displaystyle x\) be the number of dimes Julie has and \(\displaystyle y\) be the numbers of quarters she has. The number of dimes and the number of quarters add up to \(\displaystyle 35\) coins. The value of all quarters and dimes is \(\displaystyle \$7.25\). We can then write the following system of equations:

\(\displaystyle x+y=35\)

\(\displaystyle 0.1x+0.25y=7.25\)

To use substitution to solve the problem, begin by rearranging the first equation so that \(\displaystyle x\) is by itself on one side of the equals sign:

\(\displaystyle x=35-y\)

Then, we can replace \(\displaystyle x\) in the second equation with \(\displaystyle 35-y\):

\(\displaystyle 0.1(35-y)+0.25y=7.25\)

Distribute the \(\displaystyle 0.1\):

\(\displaystyle 3.5-0.1y+0.25y=7.25\)

Subtract \(\displaystyle 3.5\) from each side of the equation:

\(\displaystyle 0.15y=3.75\)

Divide each side of the equation by \(\displaystyle 0.15\):

\(\displaystyle y=25\)

Now, we can insert our value for \(\displaystyle y\) into the first equation and solve for \(\displaystyle x\):

\(\displaystyle x=35-y=35-25=10\)

Julie has \(\displaystyle 25\) quarters and \(\displaystyle 10\) dimes.

To solve a system of two equations with two unknowns, we first solve one of the equations for one of the variables and then substitute that value into the other equation. This allows us to find a solution for one of the variables, which we then plug back into either equation to find the solution for the other variable:

\(\displaystyle -3y+x=2\rightarrow x=3y+2\)

Substituting the right side of the rearranged equation into the other equation for \(\displaystyle x\), we get:

\(\displaystyle 4-y=3x\)

\(\displaystyle 4-y=3(3y+2)\)

Now we can solve this equation for \(\displaystyle y\).

\(\displaystyle 4-y=9y+6\)

\(\displaystyle -10y=2\rightarrow y=-\frac{1}{5}\)

Now that we know the value of \(\displaystyle y\), we can plug that value into the other equation for \(\displaystyle y\) and solve for \(\displaystyle x\):

\(\displaystyle x=3y+2\)

\(\displaystyle x=3(-\frac{1}{5})+2=-\frac{3}{5}+2=-\frac{3}{5}+\frac{10}{5}=\frac{7}{5}\)

We're told to find the slope and \(\displaystyle y\)-intercept of a line that passes through the points \(\displaystyle (-6,4)\) and \(\displaystyle (14,6)\). To begin, calculate the slope using the following equation:

\(\displaystyle \small m=\frac{y-y'}{x-x'}=\frac{6-4}{14--6}=\frac{2}{20}=\frac{1}{10}\)

So now that we have our slope, we need to find our \(\displaystyle y\)-intercept.

Recall the general form for a linear equation:

\(\displaystyle \small y=mx+b\)

Rearrange to solve for \(\displaystyle b\) and use our slope and one of the given points to solve:

\(\displaystyle \small \small y-mx=b \rightarrow b=y-mx\)

\(\displaystyle \small b=6-14*\frac{1}{10}{}=6-1.4=4.6\)

So, we have our slope, \(\displaystyle \frac{1}{10}\) and our \(\displaystyle y\)-intercept, \(\displaystyle 4.6\).

We can solve this problem by setting up a system of equations and using elimination:

\(\displaystyle 5x-7y=-2\)

\(\displaystyle 8y-x=7\)

We can eliminate the \(\displaystyle x\) and solve for \(\displaystyle y\) by multiplying the bottom equation by \(\displaystyle 5\) and adding the equations:

   \(\displaystyle 5x-7y=-2\)

\(\displaystyle -5x+40y=35\)

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               \(\displaystyle 33y=33\)

                    \(\displaystyle y=1\)

We can now find \(\displaystyle x\) by substituting our \(\displaystyle y\) into any equation:

\(\displaystyle 8(1)-x=7\)

\(\displaystyle 8-7=x\)

\(\displaystyle x=1\)

We have enough information to write out two equations:

\(\displaystyle n\cdot p=24\)

\(\displaystyle n+p=11\)

Using the first equation, we can narrow our potential values to:\(\displaystyle \textup{ 1 and 24, 2 and 12, 3 and 8, and 4 and 6}\).

Using the second equation, we can narrow down our values even further to \(\displaystyle \textup{3 and 8}\).  We are, however, being asked specifically for the value of \(\displaystyle p\). Since we cannot state if the \(\displaystyle 3\) or the \(\displaystyle 8\) represents \(\displaystyle p\) and which represents \(\displaystyle n\), we cannot answer this question. Additional data, such as \(\displaystyle p\) is less than \(\displaystyle 5\), would be required.

We can solve this problem in the same way we would solve a system of equations using elimination. Since we are solving for \(\displaystyle x\) we can manipulate the system to cancel out the \(\displaystyle y\) values:     \(\displaystyle 3(3x-2y=16)\)                                        \(\displaystyle 9x-6y=48\)

 \(\displaystyle 2(-2x+3y=-4)\)                                \(\displaystyle -4x+6y=-8\)

We then add the equations. Notice how the \(\displaystyle y\) values cancel out \(\displaystyle 5x=40\)

leaving us with \(\displaystyle x=8\)