Question 279308
Let {{{a}}} = cost of a battery
Let {{{b}}} = cost of a case of spark plugs
Let {{{c}}} = cost of a dozen pair of wiper blades
given:
(1) {{{6a + 5b + 2c = 830}}}
(2) {{{3a + 7b + 4c = 820}}}
(3) {{{a = 2c - 22}}}
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This is 3 equations and 3 unknowns, so it's solvable
Multiply both sides of (2) by {{{2}}} and subtract
(1) from (2)
(2) {{{6a + 14b + 8c = 1640}}}
(1) {{{-6a - 5b - 2c = 830}}}
{{{9b + 6c = 810}}}
{{{9b = 810 - 6c}}}
{{{b = 90 - (2/3)*c}}}
By substitution:
(2) {{{3a + 7b + 4c = 820}}}
{{{3*(2c - 22) + 7*(90 - (2/3)*c) + 4c = 820}}}
{{{6c - 66 + 630 - (14/3)*c + 4c = 820}}}
{{{(16/3)*c = 820 - 630 + 66}}}
{{{(16/3)*c = 256}}}
{{{16c = 768}}}
{{{c = 48}}}
And, from (3),
{{{a = 2c - 22}}}
{{{a = 2*48 - 22}}}
{{{a = 74}}}
And, from (1)
(1) {{{6a + 5b + 2c = 830}}}
{{{6*74 + 5b + 2*48 = 830}}}
{{{444 + 5b + 96  = 830}}}
{{{5b = 830 - 444 - 96}}}
{{{5b = 290}}}
{{{b = 58}}}
$74 = cost of a battery
$58 = cost of a case of spark plugs
$48 = cost of a dozen pair of wiper blades 
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check answer:
(2) {{{3a + 7b + 4c = 820}}}
{{{3*74 + 7*58 + 4*48 = 820}}}
{{{222 + 406 + 192 = 820}}}
{{{820 = 820}}}
OK