2 Equations But 3 Unknowns?

Algebra Level 1

$\begin{array} { c c c c c c c } 3a &+& 7b &+ &c &= & 315 \\ 4a &+& 10b &+& c &= & 420 \\ a &+& b &+& c &= & ? \end{array}$

105 100 110 150

9 solutions

Alan Enrique Ontiveros Salazar
May 6, 2016

Multiply the first equation by 3: $9a+21b+3c=945$ and multiply the second equation by 2: $8a+20b+2c=840$ , subtract them: $a+b+c=945-840=\boxed{105}$ .

Moderator note:

Here's a systematic way of understanding why Alan chose to multiply each equations by a certain scalar multiple:

Consider multiplying the first equation by $m$ and the second equation by $n$ ,

$3ma + 7mb + mc = 315m, \quad 4na + 10nb + nc = 420n .$

Taking their difference gives $a(3m-4n) + b(7m - 10n) + c(m -n) = 315 - 420n$ . We want to find the value of $m$ and $n$ satisfying $3m-4n = 7m - 10n = m-n$ . Solving these equations simultaneously gives $m=3,n=2$ .

Another way to solve question is by using kernels. We can rewrite the 2 given equations as the matrix, $\begin{bmatrix}{3} && {7} && {1} \\ {4} && {10} && {1}\end{bmatrix}$ .

The kernel of this matrix consists of all vectors $( a,b,c)$ for which

$\begin{bmatrix}{3} && {7} && {1} \\ {4} && {10} && {1}\end{bmatrix} \begin{bmatrix}{a} \\ {b} \\ c \end{bmatrix} = \begin{bmatrix}{0} \\ {0} \end{bmatrix} .$

Which can be written in matrix as

$\left [ \begin{array} { c c c | c } 3 & 7 & 1 & 315 \\ 4 & 10 & 1 & 420 \\ \end{array} \right ] \stackrel{R_1\ce{->} 4R_1 , R_2 \ce{->} 3R_2 }{\huge \longrightarrow} \left [ \begin{array} { c c c | c } 12 & 28 & 4 & 1260 \\ 12 & 30 & 3 & 1260 \\ \end{array} \right ] \stackrel{R_2 \ce{->} R_2 - R_1}{\huge \longrightarrow } \left [ \begin{array} { c c c | c } 12 & 28 & 4 & 1260 \\ 0 & 2 & -1 & 0 \\ \end{array} \right ]$

This tells us that $2y = z$ . By introducing a dummy variable $t$ , we can write $(y,z) = (2t,t )$ . Substituting these values into $3x+7y + z = 315$ gives $x = 105 - 3t$ . Hence,

$x + y + z = (105 - 3t) + (2t ) + (t) = \boxed{105} .$

Good solution.

Chew-Seong Cheong - 5 years, 1 month ago

Thank you Alan for the solution and thank you to challenge master for the systematic way

Pil Pinas - 4 years, 11 months ago

Chew-Seong Cheong
May 7, 2016

It is given that: $\begin{cases} 3a + 7b + c = 315 & ...(1) \\ 4a + 10b + c = 420 &...(2) \\ a+b+c = ? &...(3) \end{cases}$

Then, we have:

$\begin{aligned} (2)-(1): \quad \quad \quad \quad \, \, a + 3b & = 105 \\ \implies \color{#3D99F6}{a} & = \color{#3D99F6}{105 - 3b} \\ (1): \quad 3(\color{#3D99F6}{105 - 3b}) + 7b + c & = 315 \\ 315 -9b + 7b + c & = 315 \\ -2b + c & = 0 \\ \implies \color{#D61F06}{c} & = \color{#D61F06}{2b} \end{aligned}$

Now,

$\begin{aligned} (3): \quad \color{#3D99F6}{a} + b + \color{#D61F06}{c} & = \color{#3D99F6}{105 - 3b} + b + \color{#D61F06}{2b} \\ & = \boxed{105} \end{aligned}$

Emanuel Dicker
May 3, 2016

Given:
$3a + 7b + c = 315$
$4a + 10b + c = 420$

$4(3a + 7b) = 4(315 -c)$
$+ -3(4a + 10b) = -3(420 - c)$
$=$
$12a -12a +28b - 30b = 1260-1260 -4c + 12c =$
$-2b = 8c$
~~ $c= -.25b$

$4a + 10b + c = 420$
$- 3a + 7b + c = 315$
$= a + 3b = 105$
~~ $3b = 105- a$
$b = 35- 1/3a$

$4a + 10b + c = 420$
$- a + 3b= 105$
$= 3a + 7b = 315$
$3a + 7(35-1/3a) = 315$
$2/3a = 70$
$a = 105$

$a + 3b = 105$
$105 + 3b = 105$
$3b = 0$
$b= 0$

$4a + 10b +c = 420$

$4(105) + 10(0) + c = 420$
$420 + c = 420$
$c = 0$

Thus, $a + b + c = 105$

Nice answer

Fidel Simanjuntak - 5 years, 1 month ago

In Response to Fidel Simanjuntak: Thanks!

Emanuel Dicker - 5 years, 1 month ago

Zain Majumder
Oct 19, 2017

After subtracting equation 1 from equation 2, we get:

$a+3b=105$

$2a+6b=210$

Expand the first equation to get:

$2a+6b+a+b+c=315$

$210+a+b+c=315$

$a+b+c= \boxed{105}$ .

Kamlesh Tiwari
May 19, 2016

3a + 7b +c = 315 --- (1)

4a + 10b + c = 420 --- (2)

a + b + c = ? --- (3)

Now, Eq (2) - Eq (1) => a + 3b = 105
Multiply this equation by 3

                   => 3a + 9b = 315       --(4)

Now, Eq (2) - Eq (4) : 4a + 10b + c - 3a - 9b = 420 - 315

                            => a + b + c = 105

Amed Lolo
May 18, 2016

eq#(2)-eq#(1)=4a+10b+c-3a-7b-c=420-315. =105,,,,so a+3b=105,,,a+b+c=U(eq#3). eq#1-eq#3,,,2a+6b=315-U=2(a+3b). 315-U=2×105,,,,,SO. U=315-210=105######

谦艺伍
May 13, 2016

Let (1) denote the first equation and (2) denote the second equation.

(2) - (1): $(4a+10b+c)-(3a+7b+c) = 420-315$ $a+3b=105......(3)$ $(1) - 2\times(3):$ $(3a+7b+c)-2(a+3b) = 315-2(105)$ $a+b+c=105$

Ramiel To-ong
May 11, 2016

nice solution.

Sabhrant Sachan
May 5, 2016

$\text {From 1st Equation } 3a+7b+c=315 \\ \implies c=315-7b-3a \\ \text {Subtract 1st and 2nd Equation , we get : } \\ \implies a+3b=105 \\ \text {Put Value of c in equation 3 } \\ \implies a+b+315-3a-7b \\ \implies -2(a+3b)+315 \\ \implies -2\times105+315 \\ \color{#3D99F6}{\boxed{\text{Ans = } 105 }}$

2 Equations But 3 Unknowns?

9 solutions

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