Simultaneous equations (Systems of linear equations): Difficult Problems with Solutions

Solution:
[tex]\begin{cases} x-y=3 \\ x^2-y^2=33 \end{cases}[/tex]
Use the difference of squares: [tex]x^2-y^2=(x-y)(x+y)=3(x+y)=33 \Rightarrow x+y=11[/tex]
[tex]\begin{cases} x-y=3 \\ x+y=11 \end{cases} \Rightarrow x=7,\ y=4[/tex]
The numbers are [tex]7[/tex] and [tex]4[/tex].

Solution:
Expand the brackets; the terms with [tex]xy[/tex] cancel:
[tex](xy+2x-y-2)-(xy+5x-2y-10)=-3x+y+8=0[/tex], so [tex]y=3x-8[/tex].
[tex](xy-3x+4y-12)-(xy-4x+7y-28)=x-3y+16=0[/tex]
Substitute into the second equation: [tex]x-3(3x-8)+16=0 \Rightarrow -8x+40=0 \Rightarrow x=5[/tex], [tex]y=3\times 5-8=7[/tex].
The solution of the system is [tex](5, 7)[/tex].

Solution:
Introduce new unknowns [tex]u=x+y[/tex], [tex]v=x-y[/tex]:
[tex]\begin{cases} 5u-7v=2 \\ 2u-3v=4 \end{cases}[/tex]
Multiply the first equation by [tex]2[/tex] and the second by [tex]5[/tex]: [tex]10u-14v=4[/tex], [tex]10u-15v=20[/tex].
Subtracting the second equation from the first, we get [tex]v=-16[/tex]. [tex]5u=2+7(-16)=-110 \Rightarrow u=-22[/tex]
[tex]x=\frac{u+v}{2}=\frac{-22-16}{2}=-19[/tex], [tex]y=\frac{u-v}{2}=\frac{-22+16}{2}=-3[/tex]
The solution of the system is [tex](-19, -3)[/tex].

Solution:
Introduce new unknowns [tex]u=\frac{1}{x}[/tex], [tex]v=\frac{1}{y}[/tex] ([tex]x\neq 0[/tex], [tex]y\neq 0[/tex]):
[tex]\begin{cases} 2u+v=\frac{1}{6} \\ u-3v=\frac{11}{6} \end{cases}[/tex]
From the second equation [tex]u=\frac{11}{6}+3v[/tex]: [tex]\frac{11}{3}+6v+v=\frac{1}{6} \Rightarrow 7v=-\frac{7}{2} \Rightarrow v=-\frac{1}{2}[/tex]
[tex]u=\frac{11}{6}-\frac{3}{2}=\frac{1}{3}[/tex]
[tex]x=\frac{1}{u}=3[/tex], [tex]y=\frac{1}{v}=-2[/tex]
The solution of the system is [tex](3, -2)[/tex].

Solution:
Multiply the first equation by [tex]\sqrt{2}[/tex] and the second by [tex]\sqrt{3}[/tex]:
[tex]\begin{cases} -\sqrt{6}\,x+4y=2\sqrt{3} \\ \sqrt{6}\,x-3y=-\sqrt{3} \end{cases}[/tex]
Adding the equations, we get [tex]y=\sqrt{3}[/tex].
[tex]\sqrt{2}\,x-\sqrt{3}\times \sqrt{3}=-1 \Rightarrow \sqrt{2}\,x=2 \Rightarrow x=\frac{2}{\sqrt{2}}=\sqrt{2}[/tex]
The solution of the system is [tex]\left(\sqrt{2}, \sqrt{3}\right)[/tex].

Solution:
A system [tex]\begin{cases} a_1x+b_1y=c_1 \\ a_2x+b_2y=c_2 \end{cases}[/tex] has no solution when [tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}[/tex] (the lines are parallel).
[tex]\frac{3}{b}=\frac{a}{6}\neq\frac{1}{1}[/tex]
[tex]\frac{3}{b}=\frac{a}{6} \Rightarrow ab=18[/tex]; [tex]\frac{a}{6}\neq 1 \Rightarrow a\neq 6[/tex]
If [tex]a=6[/tex] (and [tex]b=3[/tex]), the two equations coincide and the system has infinitely many solutions.

Solution:
Let the father be [tex]x[/tex] years old and the son [tex]y[/tex] years old now:
[tex]\begin{cases} x-3=7(y-3) \\ x+5=3(y+5) \end{cases} \Rightarrow \begin{cases} x=7y-18 \\ x=3y+10 \end{cases}[/tex]
[tex]7y-18=3y+10 \Rightarrow 4y=28 \Rightarrow y=7[/tex], [tex]x=31[/tex]
The father is [tex]31[/tex] years old and the son is [tex]7[/tex]. Check: [tex]28=7\times 4[/tex] and [tex]36=3\times 12[/tex].

Solution:
Let [tex]AB=a[/tex] and [tex]BC=b[/tex]:
[tex]\begin{cases} (a-2)(b+1)=ab \\ (a+3)(b-1)=ab \end{cases} \Rightarrow \begin{cases} a-2b-2=0 \\ -a+3b-3=0 \end{cases}[/tex]
Adding the equations, we get [tex]b-5=0 \Rightarrow b=5[/tex], [tex]a=2b+2=12[/tex].
[tex]AB=12[/tex] in, [tex]BC=5[/tex] in

Solution:
Let the number be [tex]10a+b[/tex], where [tex]a[/tex] is the tens digit and [tex]b[/tex] the units digit:
[tex]\begin{cases} 10a+b=9b+4 \\ 10a+b=11a+1 \end{cases} \Rightarrow \begin{cases} 10a-8b=4 \\ b=a+1 \end{cases}[/tex]
[tex]10a-8(a+1)=4 \Rightarrow 2a=12 \Rightarrow a=6[/tex], [tex]b=7[/tex]
The number is [tex]67[/tex]. Check: [tex]67=9\times 7+4[/tex], [tex]67=11\times 6+1[/tex].

Solution:
[tex]\begin{cases} a+b+c=1 \\ 4a+2b+c=0 \\ 9a+3b+c=3 \end{cases}[/tex]
Subtract the first equation from the second and the second from the third:
[tex]3a+b=-1[/tex], [tex]5a+b=3 \Rightarrow 2a=4 \Rightarrow a=2,\ b=-7[/tex]
[tex]c=1-a-b=1-2+7=6[/tex]
[tex]f(0)=c=6[/tex]

Solution:
Add all four equations — each unknown appears three times:
[tex]3(x+m+t+p)=510+680+770+620=2580 \Rightarrow x+m+t+p=860[/tex]
Subtract each equation from this sum:
[tex]p=860-510=350[/tex], [tex]x=860-680=180[/tex], [tex]m=860-770=90[/tex], [tex]t=860-620=240[/tex]

Solution:
Eliminate [tex]z[/tex]: subtract the second equation from the first, and add the second equation multiplied by [tex]2[/tex] to the third:
[tex]\begin{cases} x-4y=-7 \\ 5x+3y=11 \end{cases}[/tex]
[tex]x=4y-7 \Rightarrow 5(4y-7)+3y=11 \Rightarrow 23y=46 \Rightarrow y=2[/tex], [tex]x=1[/tex]
[tex]z=6-x-y=3[/tex]
The solution of the system is [tex](1, 2, 3)[/tex].

Solution:
Introduce new unknowns [tex]u=|x+1|\geq 0[/tex], [tex]v=|y-1|\geq 0[/tex]:
[tex]\begin{cases} u+v=7 \\ u-v=1 \end{cases} \Rightarrow u=4,\ v=3[/tex]
[tex]|x+1|=4 \Rightarrow x=3[/tex] or [tex]x=-5[/tex]
[tex]|y-1|=3 \Rightarrow y=4[/tex] or [tex]y=-2[/tex]
Every value of [tex]x[/tex] can be combined with every value of [tex]y[/tex], so the system has [tex]4[/tex] solutions:
[tex](3, 4)[/tex], [tex](3, -2)[/tex], [tex](-5, 4)[/tex], [tex](-5, -2)[/tex]
A common mistake is to substitute [tex]x=-5[/tex] into the second equation instead of [tex]|x+1|=4[/tex]: the equation contains [tex]|x+1|[/tex], not [tex]x[/tex].