I tried to solve these systems of equations in my book: \begin{align} 7&x+4y+3z+2w=46\\ 5&x-y+4w=23\\ &x+z=6\\ 3&x+7w=15 \end{align}
I tried to solve it in many different ways, but I still haven't gotten $x=5$, $y=2$, $z=1$ and $w=0$ which are the solutions of these systems of equations.
Did I make a mistake? How did the book get $x=5$, $y=2$, $z=1$ and $w=0$?
With $z = 6 - x$ we can eliminate $z$ so we get:
$$7x+4y+3(6-x)+2w=46$$
Then distributing the $3$: $$7x+4y+18-3x+2w=46$$
Moving variable terms to left of the left side and constant term to right of left side then combining like terms we get: $$4x+4y+2w+18=46$$
Subtracting $18$ from both sides so we get $$4x+4y+2w=28$$
Dividing by $2$, then it becomes: $$2x+2y+w=28$$ $$5x-y+4w=23$$ $$3x+7w=15$$
Then deriving from the first equation we get
$$w=14-2x-2y$$
Substituting $w$ we get:
$$5x-y+4(14-2x-2y)=23$$ $$3x+7(14-2x-2y)=15$$
By the process the steps for the first one are below:
$$5x-y+56-8x-8y=23$$ $$5x-y-8x-8y+56=23$$ $$-3x-9y+56=23$$ $$-3x-9y=-33$$
Let's get rid of the negatives:
$$3x+9y=33$$
And the steps for the second one are below:
$$3x+98-14x-14y=15$$ $$-11x-14y+98=15$$ $$-11x-14y=-83$$
Get rid of the negatives, there are the 2 equations:
$$3x+9y=33$$ $$11x+14y=83$$
We could do elimination here:
$$11(3x+9y)=11\times 33$$ $$3(11x+14y)=3\times 83$$
We get then:
$$33x+99y=363$$ $$33x+42y=249$$
Subtracting one from the other we get:
$$(33x+99y)-(33x+42y)=363-249$$
We can distribute the minus sign on the left side:
$$33x+99y-33x-42y=363-249$$
Simplifying we get:
$$57y=114$$
Dividing both sides by $57$ we get:
$$y=2$$
Now let's plug this in into any one of the equations before multiplying (in this case, I used $3x+9y=33$ just for simplicity, simplifying, through the steps:
$$3x+9\times 2=33$$ $$3x+18=33$$ $$3x=15$$ $$x=5$$
We've got $x=5$ and $y=2$. Can you finish the rest to get $z=1$ and $w=0$?